Lie Groups Fall 2016 (Cohen) Lecture Notes

Home , Identity component, Lie group, Simple Lie algebra, Simple Lie group, Solvable Lie algebra

LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

1. Preliminaries Remark 1.1. The content in these notes is Frankensteined together from many sources, including Knapp’s Lie Groups Beyond An Introduction, Bump’s Lie Groups, Tao’s Hilbert’s Fifth Problem and Related Topics, Varadarajan’s Lie Groups, Lie Algebras, and Their Representations, Hilgert and Neeb’s Structure and Geometry of Lie Groups, and the author’s own paltry brain. Although I have added details to many proofs or changed them entirely, I am not making any serious attempt to avoid plagiarism and primary credit belongs to those authors. In particular many of the exercises appear in Knapp. 1.1. Topological Groups. Deﬁnition 1.2. A topological group is a pair (G, τ) where G is a group, and τ is a topology on G satisfying the following conditions: (1) the mapping (g, h) 7→ gh, G × G → G is continuous (with respect to τ); and (2) the mapping g 7→ g−1, G → G is continuous (with respect to τ).

We will always refer to the identity element of G as eG, or, if the context is clear, just as e. Remark 1.3. Most of the time the particular topology τ will be assumed or understood in context, and we will just refer to G itself as a topological group (rather than the pair (G, τ)). Example 1.4 (Examples of Topological Groups). Any abstract group equipped with the discrete topology; any abstract group equipped with the trivial topology; (Z, +); (Q, +); (R, +); (Rn, +); (R+, ·); the circle group S1; GL(n, R); GL(n, C); SL(n, R); SL(n, C); any Banach space regarded as an additive group equipped with the norm topology; many homeomorphism groups equipped with the compact-open topology; .... Exercise 1. Let G be a group and τ a topology on G. Prove that (G, τ) is a topological group iﬀ the mapping (g, h) 7→ gh−1, G × G → G is continuous with respect to τ. Exercise 2. Let G be a topological group, and let h ∈ G be arbitrary. Deﬁne the mappings i : G → G, i(g) = g−1; Lh : G → G, Lh(g) = hg; Rh : G → g, Rh(g) = gh.

Prove that i, Lh, and Rh are self-homeomorphisms of G. Deﬁnition 1.5. Let G be a group and let A, B ⊆ G. Let n be a positive integer. We deﬁne A−1 = {a−1 : a ∈ A}; AB = {ab : a ∈ A, b ∈ B}; n A = {a1a2...an : a1, a2, ..., an ∈ A}.

A is called symmetric if A = A−1. Exercise 3. Let G be a topological group and let U ⊆ G be an open neighborhood of e. Prove that there exists an open neighborhood V of e such that V 2 ⊆ U. Exercise 4. Let G be a topological group and let U ⊆ G be an open neighborhood of e. Prove that for every positive integer n, there exists an open neighborhood V of e such that V n ⊆ U. Exercise 5. Let G be a topological group and let U ⊆ G be an open neighborhood of e. Prove that for every positive integer n, there exists a symmetric open neighborhood V of e such that V n ⊆ U. 1 2 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

1.2. Diﬀerentiable Manifolds.

Definition 1.6. A smooth premanifold of dimension n (n ∈ N) is a Hausdorff topological space M together with a set U of pairs (U, φ) (called charts), where U is an open subset of M and φ : U → Rn is a mapping, satisfying (i) the collection {U : ∃φ(U, φ) ∈ U} is an open cover of M; (ii) for each (U, φ) ∈ U, φ is a homeomorphism of U onto an open subset of Rn; and (iii) for each (U, φ), (V, ψ) ∈ U the map ψ◦φ−1 is a smooth diffeomorphism from φ(U ∩V ) onto ψ(U ∩V ). The set U is called a preatlas.

If M and N are smooth premanifolds, a continuous map f : M → N is called smooth if whenever (U, φ) and (V, ψ) are charts of M and N respectively, the map ψ ◦ f ◦ φ−1 is a smooth map from φ(U ∩ f −1(V )) into ψ(V ). The smooth map f is called a diﬀeomorphism if it is a bijection and has a smooth inverse.

If M is a smooth premanifold with preatlas U, then we let U 0 denote the set of all pairs (U, φ) where U is an open subset of M and φ is a diﬀeomorphism of U onto an open subset of Rn. If U = U 0, then we call U an atlas and M a smooth manifold.

Suppose M is a smooth manifold and m ∈ M. Suppose U is an open neighborhood of m and (U, φ) ∈ U, where φ(m) is the origin in Rn. Then we may write

φ(u) = (x1(u), x2(u), ..., xn(u)) for u ∈ U where the xi’s (1 ≤ i ≤ n) are smooth functions from U to R. In this case we call the tuple (x1, .., xn) local coordinates at m or coordinate functions on U. The set U is called a coordinate patch or coordinate neighborhood. Exercise 6. Show that if M is a smooth manifold and m ∈ M, then there exist local coordinates at m. Exercise 7. Show that if f : N → P and g : M → N are smooth maps between smooth manifolds M,N,P , then the composition f ◦ g is also a smooth map from M into P . Fact 1.7. If M and N are smooth manifolds, then the product M × N can be made into a smooth premanifold in a natural way by defining the charts in M × N to be products of the charts in M and N, respectively. M ×N then admits a natural smooth manifold structure, in such a way that m 7→ (m, n) is a smooth diffeomorphism of M onto M ×{n} for each n ∈ N, and n 7→ (m, n) is a smooth diffeomorphism of N onto {m} × N. Definition 1.8. A (real) Lie group is a separable topological group G which is also a smooth manifold, satisfying the following conditions: (1) the mapping (g, h) 7→ gh, G × G → G is smooth; and (2) the mapping g 7→ g−1, G → G is smooth.

Proposition 1.9. Let G be a Lie group. The maps i, Lh, and Rh (h ∈ G) deﬁned in Exercise 2 are self-diﬀeomorphisms of G.

Proof. We already know i, Lh, and Rh are self-homeomorphisms, so we need only show they are each smooth with smooth inverse. Since i is its own inverse, it has a smooth inverse. For any h ∈ G, for every g ∈ G, we have Lh(g) = hg = ·(h, g), so Lh is the composition of the smooth map g 7→ (h, g) with smooth multiplication. Therefore Lh is smooth, and similarly for Rh. The inverses of Lh and Rh are Lh−1 and Rh−1 , respectively, which are smooth maps by the previous two sentences.

2. Classical Lie Groups and Lie Algebras 2.1. Linear Groups.

Deﬁnition 2.1. The general linear groups are the groups GL(n, R) of nonsingular n×n real matrices, and GL(n, C) of nonsingular n × n complex matrices, for n ∈ Z+. The underlying group operation for each of these groups is matrix multiplication. We also deﬁne: LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 3

SL(n, C) = {X ∈ GL(n, C) : det X = 1} SL(n, R) = {X ∈ GL(n, R) : det X = 1} t O(n) = {X ∈ GL(n, R): XX = 1} t SO(n) = {X ∈ GL(n, R): XX = 1, det X = 1} ∗ U(n) = {X ∈ GL(n, C): XX = 1} ∗ SU(n) = {X ∈ GL(n, C): XX = 1, det X = 1}

The above, respectively, are the special linear groups over C and over R, the orthogonal groups, the special orthogonal groups, the unitary groups, and the special unitary groups for n ∈ Z+.

We will always denote the identity matrix by I, or In if the context is not clear.

Remark 2.2. For each n ∈ Z+, the general linear group GL(n, C) is an open subset of the Euclidean n2 2n2 space Matn(C) of n×n complex matrices (which we identify with C or R ). So GL(n, C) is a smooth manifold in a natural way, and a Lie group under this smooth structure. Similarly, each of the other groups listed in the previous deﬁnition sits inside GL(n, C) as a closed subgroup, and becomes a smooth manifold by inheriting the structure of GL(n, C). Each group then can be viewed as a Lie group. 2.2. The Lie Algebra of a Linear Group.

Deﬁnition 2.3. Let G be a closed subgroup of GL(n, C). A smooth curve in G is a smooth mapping c : J → G, where J is a subinterval of R with nonempty interior.

The Lie algebra of G is the set

g = {c0(0) : c is a smooth curve in G with c(0) = I}.

So g is a set of matrices of the same size as G. The members of g are not necessarily invertible.

Example 2.4. If G = SO(3), then

 cos t sin t 0  c(t) =  − sin t cos t 0  for t ∈ (−1, 1) 0 0 1

is an example of a smooth curve in G with c(0) = I3. In this case

 0 1 0  c0(0) =  −1 0 0  0 0 0

is a member of the Lie algebra g = so(3).

Proposition 2.5. The Lie algebra g of a closed subgroup G of GL(n, C) is a (real) vector subspace of Matn(C). Proof. We need to check that g is closed under addition and scalar multiplication. So let X,Y ∈ g, and let c, d be smooth curves in G for which c(0) = d(0) = I, c0(0) = X and d0(0) = Y . Then t 7→ c(t)d(t) is d 0 0 a smooth curve in G with c(0)d(0) = I. By the chain rule, dt c(t)d(t) = c(t)d (t)+c (t)d(t), and therefore 4 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

d c(t)d(t) = c(0)d0(0) + c0(0)d(0) dt 0 = X + Y.

Therefore X + Y ∈ g and g is closed under addition. To show that g is closed under scalar multiplication, check that for any k ∈ R, the mapping t 7→ c(kt) is a smooth curve in G sending 0 to I, and whose derivative evaluated at 0 is kX.

Proposition 2.6. Let G be a closed subgroup of GL(n, C) and let g be the Lie algebra of G. For each g ∈ G, deﬁne the linear map

Ad(g)X = gXg−1.

Then g is closed under Ad(g) for every g ∈ G, and the mapping Ad : G × g → g defines a smooth action of G on g. Proof. Fix any g ∈ G and let X ∈ g. Let c be a smooth curve in G with c(0) = I, c0(0) = X. Then the mapping t 7→ gc(t)g−1 defines another smooth curve in G (since left and right translations by G are self-diffeomorphisms of G), satisfying gc(0)g−1 = I. Differentiating at t = 0, by the linearity of the derivative, we get that gc0(0)g−1 ∈ g as claimed. It is clear from the definitions that Ad defines a group action (check this) and that this action is smooth (since it just consists of matrix multiplications).

Deﬁnition 2.7. For any X,Y ∈ Matn(C), we deﬁne

[X,Y ] = XY − YX.

The mapping [·, ·] : Matn(C) × Matn(C) → Matn(C) is called the Lie bracket. Note that the Lie bracket is linear in each variable.

Proposition 2.8. Let G be a closed subgroup of GL(n, C) and let g be the Lie algebra of G. Then (1) g is closed under the Lie bracket operation; (2) [X,X] = 0 for all X ∈ g; (3) [X,Y ] = −[Y,X] for all X,Y ∈ g; and (4) the Jacobi identity holds: [[X,Y ],Z] + [[Y,Z],X] + [[Z,X],Y ] = 0 for all X,Y,Z ∈ g. Proof. For (i), let X,Y ∈ g. Let c be a smooth curve in G with c(0) = I, c0(0) = X. Then t 7→ Ad(c(t))Y = c(t)Y c(t)−1 is a smooth mapping into G, whose derivative evaluated at 0 is

1 d Ad(c(t)Y = lim [Ad(c(t))Y − Y ]. dt t=0 t→0 t

Now the left-hand side above exists, while the right-hand side is the limit of an expression involving only vector space operations on members of g. Since g is a vector subspace of a matrix space, it is closed topologically, and therefore this limit lies in g. We claim that the right-hand side is actually exactly equal to [X,Y ], and if we can show this the proof of (i) will be done.

To see this, ﬁrst note that c(t)c(t)−1 = 1, and therefore taking derivatives on both sides, we get 0 −1 d −1 c (t)c(t) + c(t)[ dt c(t) ] = 0, and therefore

d −1 −1 0 −1 dt c(t) = −c(t) c (t)c(t) . LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 5

Therefore

d d Ad(c(t))Y = [c(t)Y c(t)−1] dt dt = c0(t)Y c(t)−1 − c(t)Y c(t)−1c0(t)c(t)−1.

Plugging 0 into the above yields XY − YX, as claimed.

Claim (ii) is obvious, from which claim (iii) follows because [X,Y ] + [Y,X] = [X + Y,Y + X] = [X + Y,X + Y ] = 0. Proving claim (iv) is a straightforward computation. Definition 2.9. A real vector space g is called a Lie algebra if it is equipped with a bracket operation [·, ·]: g × g → g which is linear in each variable, satisfies [X,X] = 0 for all X ∈ g, and satisfies the Jacobi identity.

Plainly, the Lie algebra of a closed subgroup of GL(n, C) is a Lie algebra.

From now on we denote gl(n, C) = Matn(C) and gl(n, R) = Matn(R). Proposition 2.10. The Lie algebras of GL(n, C) and GL(n, R), respectively, are gl(n, C) and gl(n, R).

Proof. Let F = C or R. Clearly the Lie algebra of GL(n, F) is a subspace of gl(n, F) = Matn(F). Conversely, suppose X is an arbitrary n × n matrix with entries from F. We wish to show X is in the Lie algebra of GL(n, F). By the continuity of the determinant map det : Matn(F) → C, we may choose > 0 so small that t ∈ (−, ) implies |1 − det(I + tX)| < 1. It follows that that c(t) = I + tX, for − < t < , is a smooth curve into GL(n, F), since all matrices in the range have nonzero determinant. 0 We have c(0) = I and c (0) = X as desired. Example 2.11 (The Lie Algebra of SO(3)). Again take G = SO(3). We already saw in Example 2.4  0 1 0   0 0 1  that  −1 0 0  is a member of g = so(3). Very similar computations reveal that  0 0 0  0 0 0 −1 0 0  0 0 0  and  0 0 1  also lie in so(3). Since so(3) is a vector space, we have shown that 0 −1 0     0 a b  so(3) ⊇  −a 0 c  : a, b, c ∈ R = the set of skew-symmetric 3 × 3 matrices.  −b −c 0 

Actually equality holds above. For if c is a smooth curve in SO(3), then c(t)c(t)t = 1, whence c0(t)c(t)t + c(t)c0(t)t = 0 by the product rule. Plugging in t = 0, we get c0(0) + c0(0)t = 0, i.e. c0(0) is skew-symmetric. Therefore so(3) = {X ∈ gl(3, R): X + Xt = 0}. a z Exercise 8. Compute the Lie algebra g of G = : a > 0, z ∈ . 0 a−1 C Exercise 9. Compute the Lie algebra u(2) of U(2). (Hint: The dimension of u(2) is 4.) 2.3. The Exponential of a Matrix. Deﬁnition 2.12. If X is a complex n × n matrix, deﬁne ∞ X 1 exp X = eX = Xn. n! n=0

We show below that this sum always converges and thus the deﬁnition makes sense. Proposition 2.13. Let X,Y be complex n × n matrices. 6 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

(1) exp X converges. (2) exp X exp Y = exp(X + Y ) whenever X and Y commute. (3) exp X is invertible. (4) t 7→ exp tX is a smooth curve into GL(n, C) which maps 0 to I. d (5) dt exp tX = X exp tX. (6) det exp X = etr X . (7) X 7→ exp X, gl(n, C) → gl(n, C) is a smooth mapping. j j j X 1 X 1 X 1 Proof. (1) For any n × n matrix M, we have Xn ≤ kXnk ≤ kXkn, where the right n! n! n! n=i n=i n=i side tends to 0 and i, j → ∞. Hence the series for exp X is Cauchy, and converges. In fact it converges absolutely which justiﬁes some of the manipulations below.

(2) Since X and Y commute, we have

∞ ∞ X 1 X 1 exp X exp Y = Xr · Y s r! s! r=0 s=0 X 1 = XrY s r!s! r,s ∞ n X X XkY n−k = k!(n − k)! n=0 k=0 ∞ n X 1 X n = XkY n−k n! k n=0 k=0 ∞ X 1 = (X + Y )n = exp(X + Y ). n! n=0

(3) By (2), exp(−X) is the inverse of X.

(4) and (5) are proved by directly computing the derivative of t 7→ exp(tX) term-by-term.

(6) In case X is upper triangular, then so is exp X and so det exp X depends only on the diagonal di entries of exp X. Each i-th diagonal entry (1 ≤ i ≤ n) in exp(X) is exactly e where di is the i-th diagonal entry in X. So det exp X = etr X in this case.

For a general X, we write X = gY g−1 for complex matrices Y and g where Y is upper triangular. In this case det exp X = det exp gY g−1 = det g(exp Y )g−1 = det exp Y = etr Y = etr X . (7) follows from standard facts about term-by-term diﬀerentiation of series of functions. Exercise 10. Give an example of matrices X and Y for which exp X exp Y 6= exp(X + Y ).

Exercise 11. A matrix X is called nilpotent if there exists n ∈ N such that Xn = 0. A matrix X is called unipotent if X − I is nilpotent. Prove that exp is a bijection from nilpotent matrices onto unipotent matrices. (Hint: Consider Taylor series for the logarithm.)

Exercise 12. Prove that exp surjects gl(n, C) onto GL(n, C). (Hint: Consider the Jordan normal form of a given matrix in GL(n, C).) Exercise 13. Prove that exp is never surjective as a map from gl(n, R) to GL(n, R). Lemma 2.14. There exists an open neighborhood U containing the origin in gl(n, C) and an open neighborhood V of 1 in GL(n, C) such that exp : U → V is a diﬀeomorphism of U onto V . LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 7

Proof. By the previous proposition, exp is a smooth function. By the Inverse Function Theorem, it is enough to prove that the derivative matrix at X = 0 of X 7→ exp X has nonzero determinant. To see 2 this, let {Ei : 1 ≤ i ≤ 2n } be the standard basis of gl(n, C) over R, and let xi denote the corresponding coordinate and xi the corresponding coordinate function. Compute partial derivatives:

i ∂x (exp X) d i = x (exp tEj) ∂xj X=0 dt t=0 i = [x (Ej exp(tEj))]t=0

= xi(Ej) = δij.

So the derivative matrix at 0 is the identity. Corollary 2.15. There exists an open neighborhood U containing the origin in gl(n, C) and an open neighborhood V of I in GL(n, C) such that exp : U → V is a diﬀeomorphism of U onto V . Proof. We know that exp is a smooth function, and by the previous lemma it has nonsingular derivative matrix at X = 0. So the corollary follows as an immediate consequence of the Inverse Function Theorem. Lemma 2.16. If c(t) is a smooth curve in GL(n, C) with c(0) = I and c0(0) = X, then there exists a δ > 0 so that

t k lim c = exp tX k→∞ k for all t ∈ (−δ, δ).

Proof. By the previous corollary, choose U an open neighborhood of 0 in gl(n, C) and V an open neighborhood of I in GL(n, C) so that exp is a diﬀeomorphism of U onto V . Let δ > 0 be so small that c(t) is in V for all t ∈ (−δ, δ). Deﬁne

Z(t) = exp−1 c(t) for t ∈ (−δ, δ).

So Z is a smooth curve into gl(n, C) with Z(0) = 0. By the chain rule we have

Z0(0) = (exp0(0))−1c0(0) = X.

By Taylor’s formula, we get that

Z(t) = tX + O(t2),

where O(t2) is a term that is bounded for |t| < δ and remains bounded when divided by t2. Replacing t by t/k and holding t ﬁxed gives

t t t2 t 1 Z k = k X + O k2 = k X + O k2

and therefore

t 1 kZ k = tX + O k for |t| < δ. 8 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

t k t k t 1 Therefore c( k ) = (exp Z( k )) = exp kZ( k ) = exp(tX + O( k )). Letting k → ∞, we obtain the conclusion of the lemma for |t| < δ. Theorem 2.17. Let G be a closed subgroup of GL(n, C) and let g denote the Lie algebra of G. For all X ∈ g, we have exp X ∈ G. Consequently, g = {X ∈ gl(n, C) : exp tX ∈ G∀t ∈ R}. Proof. Let X ∈ g and let c be a smooth curve in G with c(0) = I, c0(0) = X. Then for all t in the t k domain of c, we have c( k ) ∈ G. Therefore the limit as k → ∞ is in G, if this limit exists, because G is topologically closed. But this limit exists by the previous lemma, and is equal to exp tX. Thus t exp tX ∈ G for all suﬃciently small t. For large t, we may choose N so large that N is in the domain t n of c, in which case we see that exp tX = (exp n X) ∈ G. So exp tX ∈ G for all t. Lemma 2.18. Let a and b be real vector subspaces of gl(n, C) with gl(n, C) = a ⊕ b. Then there exist open balls U1 about 0 in a and U2 about 0 in b, and an open neighborhood V of I in GL(n, C), so that (A, B) 7→ exp A exp B is a diﬀeomorphism of U1 × U2 onto V .

Proof. Choose bases {X1, ..., Xr} and {Y1, ..., Ys} of a and b respectively, and consider the map

 r !  s  −1 X X (u1, ..., ur, v1, ..., v2) 7→ exp  exp uiXi exp vjYj i=1 j=1

deﬁned in some open neighborhood of 0 in Rr+s, with the result written out as a linear combination of X1, ..., Xr,Y1, ..., Ys. To prove the lemma, it suﬃces to show that this mapping is locally invertible, since exp is locally invertible. So we wish to show that this mapping has nonsingular derivative matrix at 0, and apply the Inverse Function Theorem.

To compute the derivative matrix at 0, we compute the partial derivatives separately. For each partial derivative, we can set all variables but one equal to 0 before differentiating. By examination, we get that the derivative matrix at 0 is I which is nonsingular, finishing the argument. Theorem 2.19. Let G be a closed subgroup of GL(n, C) and let g denote the Lie algebra of G. Then there exist open neighborhoods U of 0 in g and V of 1 in G such that exp is a diffeomorphism of U onto V .

Proof. Choose a vector subspace s so that gl(n, C) = g ⊕ s, and apply the previous lemma to obtain open balls U1 about 0 in g and U2 about 0 in s, and an open neighborhood V of 1 in GL(n, C) such that (X,Y ) 7→ exp X exp Y is a diﬀeomorphism of U1 × U2 onto V . Let be the radius of U2. For each 1 integer k ≥ 1, form the set k+1 U2.

1 Claim: For some suﬃciently large k, if X ∈ U1 and Y ∈ k+1 U2, then exp X exp Y/∈ G unless Y = 0.

1 Proof of claim: Assume the contrary. So for every k ≥ 1, we can ﬁnd Xk ∈ U1 and Yk ∈ k+1 U2 such that Y 6= 0, and exp Xk exp Yk ∈ G. Since exp Xk ∈ G, it follows that exp Yk ∈ G. Since Yk 6= 0, we can choose an integer nk such that /2 ≤ nkkYkk ≤ . Passing to a subsequence if necessary, we may n assume that nkYk converges, say to Y . Then Y ∈ s and Y 6= 0. Since exp nkYk = (exp Yk) k ∈ G and G is topologically closed, we have exp Y ∈ G.

p Now let p, q be integers with q > 0. We wish to show that exp q Y ∈ G. Write nkp = mkq + rk rk nkp p rk with 0 ≤ rk ≤ q − 1. Then q Yk → 0 since Yk → 0. Also q Yk → q Y , so that exp mkYk exp q Yk = nkp p rk p exp Yk → exp Y . Since exp Yk → 1, lim exp mkYk exists and equals exp Y . Since exp mkYk = q q q k→∞ q mk p (exp Yk) ∈ G, this shows exp q Y ∈ G since G is topologically closed.

So exp tY ∈ G for all rational t. But then by the continuity of exp and the closedness of G, we get exp tY ∈ G for all t, and therefore Y ∈ g, which contradicts our choice of Y from s. This proves the claim. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 9

1 Continuing the proof of the theorem, by abuse of notation we now replace U2 with k+1 U2, where k is 1 sufficiently large to satisfy the conditions of the above claim, and we replace V with exp(U1) exp( k+1 U2). Then (X,Y ) 7→ exp X exp Y is still a diffeomorphism of U1×U2 onto V , and furthermore exp X exp Y ∈ G if and only if Y = 0. It follows that V ∩ G is an open neighborhood of I in G, such that exp is a diffeomorphism of U1 onto V ∩ G. Corollary 2.20 (Cartan’s Theorem for GL(n, C)). If G is a closed subgroup of GL(n, C) with Lie algebra −1 g, then G may be regarded as a Lie group in which each pair (gV, exp ◦Lg−1 ) is a compatible chart, and dim G = dim g. Proof. By the previous theorem, find U an open neighborhood of 0 in g and V an open neighborhood −1 −1 −1 −1 of 1 in G. Let g, h ∈ G. Then the mapping (exp ◦Lg−1 ) ◦ (exp ◦Lh−1 = exp ◦Lg−1h ◦ exp is a composition of smooth mappings and hence smooth, so our chosen charts are smoothly compatible, and therefore G is a smooth manifold. It remains then only to check that the multiplication mapping (g, h) 7→ gh, G × G → G and the inversion mapping g 7→ g−1, G × G are smooth, by referring to compatible charts. We omit the details here. Proposition 2.21. Let G be a connected topological group and let U be an open neighborhood of e ∈ G. Then G is generated by U, i.e. G is the smallest subgroup of G containing U. Proof. Replace U with U ∩ U −1, so U becomes an open symmetric neighborhood of e ∈ G. Then the [ n n n S subgroup generated by U is exactly H = U . Note each U is open, because U = u∈U n−1 uU is n≥1 a union of open sets. Therefore H is open. Now note that the complement G\H is a union of cosets of H, which are open, so G\H is open. Therefore H is clopen, which implies H = G. Corollary 2.22. If G is a closed subgroup of GL(n, C) and g is the Lie algebra of G, then the connected component G0 of e ∈ G is generated by exp g.

Proof. Being a connected component, G0 is connected and closed in G, hence closed in GL(n, C). We may ﬁnd an open neighborhood U of 0 in g and and open neighborhood V of I in G so that exp is a diﬀeomorphism of U onto V . Then V = exp U is a connected subset of G by the continuity of exp, i.e. V ⊆ G0. Therefore G0 is generated by exp g ⊇ V . Corollary 2.23. If G and H are closed subgroups of GL(n, C) with the same Lie algebra, then the identity components of G and H coincide. 2.4. Computing Lie Algebras.

Deﬁnition 2.24. We deﬁne the following vector subspaces of gl(n, C):

t o(n) = so(n) = {X ∈ gl(n, R): X + X = 0} ∗ u(n) = {X ∈ gl(n, C): X + X = 0} ∗ su(n) = {X ∈ gl(n, C): X + X = 0, tr X = 0} sl(n, R) = {X ∈ gl(n, R) : tr X = 0} sl(n, C) = {X ∈ gl(n, C) : tr X = 0}. Theorem 2.25. The Lie algebra of O(n) is so(n). Therefore the dimension of O(n) is (n2 − n)/2. Proof. Suppose X is an element of the Lie algebra of O(n). Let c(t) be a smooth curve in O(n) with c(0) = I, c0(0) = X. Since c(t)c(t)t = I, the product rule gives that c0(t)c(t)t + c(t)c0(t)t = 0, and plugging in 0 shows that X + Xt = 0, so X ∈ so(n).

Conversely, if X ∈ so(n), then for any t ∈ R, we have exp tX exp tXt = exp t(X + Xt) = exp 0 = I. So the inverse of exp tX is exp tXt = (exp tX)t, i.e. exp tX ∈ O(n). Therefore by Theorem 2.17, X is in the Lie algebra of O(n). 10 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Theorem 2.26. The Lie algebra of U(n) is u(n). Therefore the dimension of U(n) is n2.

Proof. Mimic the proof of the previous theorem word for word, but replacing the transpose operation with the conjugate transpose.

d 0 Lemma 2.27. Let c be a smooth curve in GL(n, C) with c(0) = In. Then dt det c(t) t=0 = tr c (0). Proof. Inductively, suppose the lemma holds for invertible matrices of size (n − 1) × (n − 1). For each 1 ≤ i ≤ n, let ai(t) denote the function corresponding to the ﬁrst row and i-th column of c(t). Let Mi(t) denote the (n − 1) × (n − 1) matrix whose j-th column coincides with the 2nd through nth rows of column (j + i − 1) mod n in c(t). By Laplace expansion for determinants

det c(t) = a1(t) det M2(t) − a2(t) det M3(t) + ... ± an−1(t) det Mn(t) ± a(t) det M1(t).

Taking derivatives using the product rule yields

d 0 d 0 d dt det c(t) = a1(t) det M2(t) + a1(t) dt det M2(t) − a2(t) det M3(t) + a2(t) dt det M3(t) + ... ± 0 d an(t) det M1(t) + an(t) dt det M1(t) .

To ﬁnish we want to plug t = 0 into the above. Note when we do this, we get a2(0) = a3(0) = ... = an(0) = 0 since c(0) = In, so many terms vanish. Similarly, for 2 ≤ i ≤ n, note that Mi(0) contains a column of all 0’s, since c(0) = In, and therefore det Mi(0) = 0 as well. On the other hand M2(0) = In−1. Looking at the remaining terms, and applying the inductive hypothesis, we get

d 0 d det c(t) = a1(0) det M2(0) + a1(0) det M2(t) dt t=0 dt t=0 0 0 = a1(0) + tr M2(0) = tr c0(0).

Theorem 2.28. Let F = R or C. The Lie algebra of SL(n, F) is sl(n, F). Therefore the dimension of SL(n, R) is n2 − 1 and the dimension of SL(n, C) is 2(n2 − 1).

Proof. If X is in the Lie algebra of SL(n, F), and c is a smooth curve in SL(n, F) satisfying c(0) = I, c0(0) = X. Then by Lemma 2.27 tr X = 0 and X ∈ gl(n, F). Conversely, if X ∈ SL(n, F), then for all tr tX t ∈ R we have det exp tX = e = 1 and tX ∈ SL(n, C).

Corollary 2.29. The Lie algebra of SO(n) is so(n). Therefore the dimension of SO(n) is (n2 − n)/2.

Proof. Because SO(n) = O(n) ∩ SL(n, R) and so(n) = so(n) ∩ sl(n, C).

Corollary 2.30. The Lie algebra of SU(n) is su(n). Therefore the dimension of SU(n) is n2 − 1.

Proof. Because SU(n) = U(n) ∩ SL(n, C) and su(n) = u(n) ∩ sl(n, C).

Exercise 14. Let F = R or C and ﬁx a positive integer n. Let J denote the 2n×2n matrix in block form 0 In J = . Deﬁne the group Sp(2n, F) = {X ∈ GL(2n, F : XJXt = J}, called the symplectic −In 0 group over F. Compute the Lie algebra of Sp(2n, F), and deduce the dimension of the group. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 11

3. Elementary Lie Theory 3.1. Vector Fields.

Deﬁnition 3.1. Given a smooth manifold M, let C∞(M) denote the associative algebra of all smooth real-valued functions on M. Fix any point p ∈ M. We say a linear mapping D : C∞(M) → R is a local derivation at p if

D(fg) = D(f)g(p) + f(p)D(g) for all f, g ∈ C∞(M).

The set TpM consisting of all derivations at p is called the tangent space to M at p.

Fix a set of local coordinates φ = (x1, ..., xn) on an open neighborhood U of p, with φ(p) = 0. We ∂ ∞ ∞ deﬁne mappings ∂xi : C (M) → C (U) (1 ≤ i ≤ n) by the rule

∂ h ∂ −1 i ∂xi f (w) = ∂x f ◦ φ (x) , i x=φ(w)

∞ ∂ for f ∈ C (M) and w ∈ U. The fact that each ∂xi depends on the choice of p and the choice of φ is not reﬂected in the notation, but this generally will not cause us a problem in context.

∂ ∞ It is straightforward to check that each mapping f 7→ ∂xi f (p), C (M) → R is a linear mapping, and a local derivation at p. For convenience, we will refer to this particular local derivation just using ∂ ∂ the notation ∂xi , so ∂xi ∈ TpM..

Proposition 3.2. Let M be a smooth manifold and p ∈ M. Then the tangent space TpM is a real vector space, and dim TpM = dim M.

∞ Proof. If D1 and D2 are local derivations at p, and f, g ∈ C (M) are arbitrary, then (D1 + D2)(fg) = D1(fg) + D2(fg) = D1(f)g(p) + f(p)D1(g) + D2(f)g(p) + f(p)D2(g) = (D1 + D2)(f)g(p) + f(p)(D1 + D2)(g). If D is a local derivation at p and c is a real scalar, then (cD)(fg) = c[D(f)g(p) + f(p)D(g)] = (cD)(f)g(p) + f(p)(cD)(g). So TpM is a vector space over R.

∂ To compute the dimension of TpM, we claim that the vectors ∂xi will comprise a basis for TpM, for any choice of local coordinates φ = (x1, ..., xn) on an open ball U ⊆ M with φ(p) = 0. To see this, let X be an arbitrary local derivation at p. We ﬁrst note that if X(1) = X(1 · 1) = X(1) · 1 + 1 · X(1) = 2X(1), and therefore X(1) = 0 (where 1 here denotes the constant function on M). Since X is linear, it follows that X(c) = 0 for any constant function c.

Next let f ∈ C∞(M) be arbitrary. Consider the function fˆ = f ◦ φ−1 : φ(U) → R. Using the fundamental theorem of calculus, for z = (z1, ..., zn) ∈ φ(U), we may write

Z 1 d fˆ(z) = fˆ(0) + fˆ(tz)dt 0 dt n X Z 1 ∂ = fˆ(0) + zi fˆ(tz)dt. ∂x i=1 0 i

Therefore for w ∈ U, we have

n X Z 1 ∂ f(w) = f(p) + xi(w) f (φ−1(tφ(w)))dt. ∂xi i=1 0 12 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

R 1 ∂ −1 Deﬁne gi(w) = 0 ∂xi f (φ (tφ(w)))dt. Then we have shown that f is a constant plus a sum of i ∂ products of the form x · gi, where gi(p) = ∂xi f (p). Now applying X to the above, and using the linearity of X as well as the local derivation property, we get

n X i X(f) = X(f(p)) + X(x · gi) i=1 n n X X i = 0 + X(xi)gi(p) + x (p)X(gi) i=1 i=1 n X ∂ = X(x ) f (p). i ∂xi i=1

Since f was arbitrary and the coeﬃcients X(xi) do not depend on f, we have shown that X is a linear ∂ combination of the operators ∂xi . It is easy to check that these operators are linearly independent, which completes the proof. Deﬁnition 3.3. A derivation on C∞(M) is a linear mapping D : C∞(M) → C∞(M) with the property that

D(fg) = D(f)g + fD(g) for all f, g ∈ C∞(M).

A derivation on C∞(M) is also called a vector field on M. Remark 3.4. We note that a vector field X on M (i.e. a derivation on C∞(M)) may be viewed as inducing a natural correspondence between each point p ∈ M, and a particular vector Xp in the ∞ tangent space TpM, because the mapping Xp(f) = X(f)(p), C (M) 7→ R is a local derivation at p, i.e. Xp ∈ TpM. We also note that the set of all vector fields on M comprises a real vector space in a natural way. Proposition 3.5. Let M be a manifold. The vector space of derivations on M is closed under the Lie bracket, i.e. [X,Y ] = X ◦ Y − Y ◦ X is a derivation whenever X and Y are. Proof. Given f, g ∈ C∞(M), we have

[X,Y ](fg) = X(Y (fg)) − Y (X(fg)) = X(Y (f)g + fY (g)) − Y (X(f)g + fX(g)) = X ◦ Y (f)g + Y (f)X(g) + X(f)Y (g) + fX ◦ Y (g) − Y ◦ X(f)g − X(f)Y (g) − Y (f)X(g) − fY ◦ X(g) = X ◦ Y (f)g + fX ◦ Y (g) − Y ◦ X(f)g − fY ◦ X(g) = (X ◦ Y − Y ◦ X)(f)g + f(X ◦ Y − Y ◦ X)(g) = [X,Y ](f)g + f[X,Y ](g). Definition 3.6. Let M and N be smooth manifolds and φ : M → N a smooth map. Let m ∈ M and n ∈ N be such that φ(m) = n. If X ∈ TmM, i.e. if X is a local derivation at m in M, then it is straightforward to check that the mapping f 7→ X(f ◦ φ) defines a local derivation at n in N. Denote this local derivation by dφm(X) ∈ TnN. The map dφm : TmM → TnN is called the differential of φ at m. Note that each differential of φ is by definition a linear mapping.

If φ is a diffeomorphism, then we can push a vector field X on M forward this way to obtain a vector field on N. We denote this vector field by φ∗X, defined by the rule (φ∗X)n = dφm(Xm) whenever φ(m) = n. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 13

Lemma 3.7 (Chain Rule for Diﬀerentials). Let φ : M → N and ψ : N → P be smooth maps between smooth manifolds. If m ∈ M and n = φ(m) ∈ N, then d(ψ ◦ φ)m = (dψn) ◦ (dφm).

∞ Proof. Given any vector X ∈ TmM and any function f ∈ C (P ), we have (d(ψ ◦ φ)mX)(f) = X(f ◦ ψ ◦ φ) = (dφm(X))(f ◦ ψ) = (dψn((dφm)(X)))(f). Exercise 15. Prove that if φ : M → N and ψ : N → P are diffeomorphisms of smooth manifolds, then (ψ ◦ φ)∗ = ψ∗ ◦ φ∗. Definition 3.8. Let G be a Lie group. A vector field X on G is called left-invariant if for every g, h ∈ G, we have

(dLg)hXh = Xgh.

We denote the space of all left-invariant vector fields on G by g, the Lie algebra of G. The bracket operation on g is defined by [X,Y ] = X ◦ Y − Y ◦ X for all X,Y ∈ g. It is straightforward to check that g is closed under the Lie bracket. Proposition 3.9. Let G be a Lie group and X a vector field on G. Then X is left-invariant if and only if (Lg)∗X = X for every g ∈ G.

Proof. Suppose X is left-invariant, let g ∈ G be arbitrary, and let Y = (Lg)∗X. Then for any h ∈ G, −1 −1 since g h = Lg (h), we have

Yh = [(Lg)∗X]h

= [(dLg)g−1hXg−1h]

= Xgg−1h

= Xh.

Therefore Y = X. Conversely, if Y = X, then for each h ∈ H, we have Lg(h) = gh, whence (dLg)hXh = [(Lg)∗X]gh = Ygh = Xgh, so X is left-invariant.

Proposition 3.10. Let G be a Lie group. For every vector V ∈ TeG, there exists a unique left-invariant vector ﬁeld X on G for which Xe = V .

Proof. Deﬁne X by letting Xe = V , and Xg = (dLg)eXe, for each g ∈ G. Observe that X is left-invariant because if g, h ∈ G, then applying the chain rule for diﬀerentials we get

(dLg)hXh = (dLg)h(dLh)eXe

= (dLgh)eXe

= Xgh.

Corollary 3.11. Let G be a Lie group. Then the Lie algebra g of G is vector space isomorphic to the tangent space TeG of G at identity.

Proof. Let X ∈ g. So X is a left-invariant vector ﬁeld, and thus Xe ∈ TeG. It is clear that X 7→ Xe is a linear mapping, and by the previous proposition, the mapping is onto. Corollary 3.12. If G is a Lie group and g is the Lie algebra of G, then dim g = dim G. 14 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Remark 3.13. If G is a closed subgroup of GL(n, C), then we have now defined two different Lie algebras g of G: one consisting of derivatives of smooth curves at e in G, and one consisting of left-invariant vector fields on G. Proposition 3.11 shows that we can identify the space of left-invariant vector fields on G with the tangent space at e in G, so to reconcile our two definitions, one needs to verify that the tangent space at e can be identified with the set of derivatives of smooth curves at e in G. This can be done in a straightforward way, and in fact the latter is frequently used as the definition of the tangent space. One also needs to verify that under this correspondence, the Lie bracket in the space of derivatives of smooth curves (i.e. using matrix multiplication) corresponds to the Lie bracket as we have defined it in the space of left-invariant vector fields. This turns out to be true. We omit the proofs here because we are not really concerned with the differential-topological details—mostly we want to have the closed subgroups of GL(n, C) available as natural examples of abstract Lie groups, in which we can perform computations directly on the Lie algebra. In practice we will not make a distinction between our various definitions of the Lie algebra, unless necessary for clarity. In particular we will frequently identify g with TeG, for any Lie group G. 3.2. Homomorphisms. Definition 3.14. Let g and h be Lie algebras. A mapping f : g → h is a Lie algebra homomorphism if f is a linear mapping, and f([X,Y ]g) = [f(X), f(Y )]h for every X,Y ∈ g. Definition 3.15. Let G and H be Lie groups with associated Lie algebras g and h, respectively, and let π : G → H be a smooth group homomorphism. Consider the differential dπe : TeG → TeH of π at e ∈ G. If X ∈ g, i.e. X is a left-invariant vector field on G, then by Proposition 3.10 there is a unique left-invariant vector field Y on H for which Ye = dπe(Xe). Let us denote this vector field as simply Y = dπ(X). Thus to each smooth homomorphism π : G → H, we have associated a mapping dπ : g → h.

Lemma 3.16. Let G and H be Lie groups with associated Lie algebras g and h, respectively, and let π : G → H be a smooth group homomorphism. Then for every g ∈ G and every X ∈ g, we have dπg(Xg) = (dπX)π(g). ∞ Proof. Since X is left-invariant, we have Xg = (Lg)∗Xe. Thus for any f ∈ C (H), we have

dπg(Xg)(f) = Xg(f ◦ π)

= ((Lg)∗X)e(f ◦ π)

= Xe(f ◦ π ◦ Lg)

= Xe(f ◦ Lπ(g) ◦ π)

= (dπX)e(f ◦ Lπ(g))

= ((Lπ(g))∗dπX)e(f)

= (dπX)π(g)(f). Lemma 3.17. Let G and H be Lie groups with associated Lie algebras g and h, respectively, and let π : G → H be a smooth group homomorphism. Let X ∈ g and Y ∈ h. Then Y = dπX if and only if for all f ∈ C∞(H), X(f ◦ π) = (Y f) ◦ π. Proof. First assume Y = dπX. Then for any f ∈ C∞(H), for any g ∈ G, we have by Lemma 3.16

(X(f ◦ π))(g) = Xg(f ◦ π)

= (dπg(Xg))(f)

= (dπX)π(g)(f) = (Y (f))(π(g)) = Y (f) ◦ π(g). LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 15

Since g was arbitrary, this shows X(f ◦ π) = (Y f) ◦ π.

To prove the converse, assume X(f ◦ π)(g) = (Y f) ◦ π(g) for any g ∈ G and any f ∈ C∞(H). Apply this equality in particular to g = e. We get dπXe(f) = Ye(f), whence dπXe = Ye. Since dπX and Y are left-invariant vector ﬁelds which agree at e, we have dπX = Y . Theorem 3.18. Let G and H be Lie groups with associated Lie algebras g and h, respectively, and let π : G → H be a smooth group homomorphism. Then dπ : g → h is a Lie algebra homomorphism. Proof. We will apply Lemma 3.17 to show that dπ[X,Y ] = [dπX, dπY ], by showing that for all f ∈ C∞(H), [X,Y ](f ◦ π) = [dπX, dπY ] ◦ π. We prove this equality below, invoking Lemma 3.17 in the second and third lines:

[X,Y ](f ◦ π) = (X ◦ Y )(f ◦ π) − (Y ◦ X)(f ◦ π) = X((dπY )(f) ◦ π) − Y ((dπX)(f) ◦ π) = (dπX ◦ dπY )(f) ◦ π − (dπY ◦ dπX)(f) ◦ π = [dπX, dπY ](f) ◦ π. 3.3. The Exponential Map. Deﬁnition 3.19. Let M be a manifold and let c :(−, ) → M be a smooth curve, for some > 0. We d may regard the interval (−, ) as a smooth manifold, and we denote by dt the standard basis vector for

Tt0 (−, ), for any choice of t0 ∈ (−, ) (see Deﬁnition 3.1). We denote

0 d c (t0) = dct0 ( dt ) ∈ Tc(t0)M. Now let m ∈ M, and let X be a vector field on M. An integral curve for X in M, with initial condition m, is a smooth curve p :(−, ) → M (for some > 0) which satisfies p(0) = m, and 0 p (t0) = Xp(t0) for all t0 ∈ (−, ). Proposition 3.20. Let V be an open subset of Rn for some positive integer n, and let Y be a vector field on V . Then an integral curve for Y in V with initial condition 0 exists, and is uniquely determined. Sketch of Proof. Written out in terms of a basis for derivations on V , we have

n X ∂ Y = a (x , ..., x ) , i 1 n ∂x i=1 i

1 n where the ai’s are smooth functions. If p(t) is a smooth curve in V , let us write p(t) = (x (t), ..., x (t)) ∞ d for t ∈ (−, ). Then for any f ∈ C and any t0 ∈ (−, ), when we apply (dpt0 ( dt )) to f, we get

d f(x1(t), ..., xn(t)) = ∇f(x1(t), ..., xn(t)) · ((x1)0(t), ..., (xn)0(t) dt t=t0 t=t0 n X i 0 ∂f 1 n = (x ) (t0) (x (t), ..., x (t)) . ∂xi i=1 t=t0

On the other hand, applying Yp(t0) to f we get 16 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES n X 1 n ∂f 1 n ai(x (t0), ..., x (t0)) (x (t), ..., x (t)) . ∂xi i=1 t=t0

So in order to obtain p an integral curve, we need to satisfy the ﬁrst-order system of diﬀerential equations given by

i 0 1 n i (x ) (t0) = ai(x (t0), ..., x (t0)) for all t0, and x (0) = 0 for 1 ≤ i ≤ n.

The existence and uniqueness of a solution to the above system is a standard result in the theory of ordinary diﬀerential equations.

Corollary 3.21. Suppose that M is a smooth manifold, m ∈ M, and X is a vector ﬁeld on M. Then an integral curve for X in M with initial condition m exists, and is uniquely determined.

Proof. Let (U, φ) be a chart in M with φ(m) = 0. Then φ : U → φ(U) is a diffeomorphism, and thus pushing forward X we obtain a vector field Y = φ∗X on φ(U), an open subset of Euclidean space. By the previous proposition, there exists a unique integral curve q :(−, ) → φ(U) for Y with initial condition 0. Then define p = φ−1 ◦ q, so p is a smooth curve in M with p(0) = m. Moreover for any ∞ f ∈ C (M), for any t0 ∈ (−, ), we have

d (p0(t ))(f) = (f ◦ p) (t ) 0 dt 0 d = f ◦ φ−1 ◦ q (t ) dt 0 d = dq f ◦ φ−1 t0 dt −1 = Yq(t0) f ◦ φ = φ−1 Y (f) ∗ q(t0)

−1 = Xφ (q(t0)) (f)

= Xp(t0)(f).

Theorem 3.22. Let G be a Lie group and g its Lie algebra. There exists a map exp : g → G such that, for any X ∈ g, t 7→ exp tX is an integral curve for the left-invariant vector ﬁeld X, with domain R. Moreover, exp(s + t)X = exp sX exp tX for all s, t ∈ R. Proof. By Corollary 3.21, we know there exists an integral curve p for the left-invariant vector ﬁeld X in G, with p(0) = e, where the domain of p is some possibly small interval (a, b). Fix any s in the domain of p, and denote

1 ps(t) = p(s)p(t) for t ∈ (a, b);

2 ps(t) = p(s + t) for t ∈ (a − s, b − s).

1 ∞ Note that ps is actually an integral curve for X because X is left-invariant: we have for any f ∈ C (G) and t0 ∈ (a, b), LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 17

d ((p1)0(t ))(f) = (f ◦ p1) (t ) s 0 dt s 0 d = (f ◦ L ◦ p) (t ) dt p(s) 0 d = dp (f ◦ L ) t0 dt p(s)

= Xp(t0)(f ◦ Lp(s))

= ((Lp(s))∗Xp(t0))(f)

= Xp(s)p(t0)(f)

= X 1 (f). ps(t0)

2 Also, ps is an integral curve for X by the chain rule:

d ((p2)0(t ))(f) = (f ◦ p2) (t ) s 0 dt s 0 d d = (f ◦ p) (s + t ) · (s + t) (t ) dt 0 dt 0 0 = (p (s + t0))(f)

= Xp(s+t0)(f)

= X 2 (f). ps(t0)

1 2 Both curves satisfy the same initial condition ps(0) = ps(0) = p(s). Therefore by uniqueness of integral curves, we see that p(s)p(t) = p(s + t), for all s, t suﬃciently near 0.

Next, we show that if p :(−a, a) → G is an integral curve for X, then we may extend the domain of p to all of R. It suffices to show that we may extend the domain to (−3a/2, 3a/2). We extend it by the rule p(t) = p(a/2)p(t − a/2) whenever a/2 < t < 3a/2, and p(t) = p(−a/2)p(t + a/2) whenever −3a/2 < t < a/2. It follows from the results of the previous paragraph that for sufficiently small a, these definitions are consistent on the regions of overlap. Moreover, by computing the derivative very similarly 1 0 to how we did for ps above (plus one application of the chain rule), we verify that p (t0) = Xp(t0) for all t0 ∈ (−3a/2, 3a/2), and thus the extended version of p is still an integral curve for X. This allows us to iterate the argument and extend the domain of p to all of R.

So we have shown that for every left-invariant vector ﬁeld X on G, there is an integral curve pX : R → G satisfying pX (s + t) = pX (s)pX (t) for all s, t ∈ R. We deﬁne exp : g → G by the rule exp X = pX (1). We note that for any s ∈ R, the map t 7→ pX (st) is an integral curve for sX, and therefore exp sX = pX (s). Thus exp(s + t)X = exp sX exp tX for all s, t ∈ R, which proves the theorem.

Proposition 3.23. Let exp now denote the matrix exponential exp : gl(n, C) → GL(n, C). Let XI ∈ gl(n, C) and let X be the unique left-invariant vector field on GL(n, C) whose vector at I is XI . Then 0 the curve p(t) = exp tXI satisfies p (t) = Xp(t) for all t ∈ R, and consequently the matrix exponential coincides with the exponential map produced in the previous theorem. (Here we identify XI and X in the natural way by identifying gl(n, C) with TI GL(n, C) with the space of left-invariant vector fields on GL(n, C); see Remark 3.13.)

Proof. First compute X. For any g ∈ GL(n, C), by left-invariance we have Xg = (Lg)∗XI = gXI . 0 0 Meanwhile p (t) = XI exp tXI = XI p(t) = p(t)XI (since XI and exp tXI commute). Therefore p (t) = Xp(t) as promised. 18 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Lemma 3.24. Let M, N be smooth manifolds and let φ : M → N be a smooth mapping. Let m ∈ M, 0 and X ∈ TmM. Let c :(−, ) → M be a smooth curve with c(0) = m and c (0) = X. Deﬁne p = φ ◦ c. 0 Then dφm(X) = p (0). Proof. We have for any f ∈ C∞(N),

d p0(0)(f) = (f ◦ φ ◦ c) (0) dt = (c0(0))(f ◦ φ) = X(f ◦ φ)

= (dφm(X))(f). Theorem 3.25 (Lie’s First Theorem). Let G be a Lie group and g its Lie algebra. Then the exponential mapping exp : g → G is smooth. Furthermore, there is an open neighborhood U of 0 in g and and open neighborhood V of e in G such that exp : U → V is a diffeomorphism. Sketch of Proof. Go back to the proofs of Proposition 3.20 and Corollary 3.21. Both proofs relied on a standard result about the existence and uniqueness of solutions to a certain system of ordinary differential equations. In fact the standard theory of ordinary differential equations says a bit more: if the coefficients ai are allowed to vary smoothly, then the solution functions xi will vary smoothly as well (as a function of the ai’s). Translating this into the language of smooth manifolds and vector fields, one deduces that the integral curve t 7→ exp tX varies smoothly as X varies smoothly, and hence exp is a smooth function on g.

Once we have that exp is smooth, we look to compute the differential (d exp)0 : T0g → TeG, and apply the inverse function theorem. We canonically identify T0g with TeG, and consider Xe ∈ T0g = TeG. Let X denote the unique left invariant vector field corresponding to Xe. Let c(t) = tXe, so c is a smooth 0 curve in g with c(0) = 0 and c (0) = Xe. Set p = exp ◦c, so p is exactly the integral curve for X in G ∞ 0 with initial condition e. Then for any f ∈ C (G), by Lemma 3.24 we have ((d exp)0Xe) = p (0) = Xe (since p is an integral curve). This shows d exp is in fact the identity map on TeG, which has non-zero Jacobian. Now the inverse function theorem tells us that exp is a local diffeomorphism.

Corollary 3.26. If G is a Lie group and g its Lie algebra, then the diﬀerential d exp : TeG → TeG is the identity. Corollary 3.27. If G is a Lie group and g its Lie algebra, then exp G generates the connected component of identity G0 in G. Theorem 3.28. Let G, H be Lie groups and let π : G → H be a smooth homomorphism. Then π ◦ exp = exp ◦dπ. Proof. Let X ∈ g and let p be the unique integral curve for X in G with initial condition e. Consider ∞ the curve q = π ◦ p in H. We have for any f ∈ C (H), for any t0 ∈ R,

d (q0(t ))(f) = (f ◦ q) 0 dt t=t0 d = (f ◦ π ◦ p) dt t=t0 0 = (p (t0))(f ◦ π)

= Xp(t0)(f ◦ π)

= (dπp(t0)Xp(t0))(f)

= (dπX)q(t0)(f). LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 19

Therefore q is the unique integral curve for dπX in H with initial condition e. Therefore π(exp X) = exp(dπX).

Corollary 3.29. If π1 and π2 are smooth homomorphisms between two Lie groups G and H such that dπ1 = dπ2, then π1 = π2 on the connected component of identity G0.

Proof. By Theorem 3.28, π1◦exp = exp ◦dπ1 = exp ◦dπ2 = π2◦exp. Since π1 and π2 are homomorphisms, this implies π1 and π2 coincide on G0 by Corollary 3.27. Corollary 3.30. Let G, H be Lie groups and let π : G → H be a smooth homomorphism. Then

(1) if dπ is onto then H0 ⊆ ran(π); (2) if dπ is one-to-one then π is one-to-one in a neighborhood of e ∈ G; and (3) if dπ is an isomorphism, then there exist neighborhoods of identity U ⊆ G and V ⊆ H such that π : U → V is a diﬀeomorphism.

Proof. If dπ is onto then H0 = hexp hi = hexp ◦dπgi = hπ ◦ exp gi = π(hexp gi) ⊆ ran(π).

If dπ is one-to-one, then exp ◦dπ is one-to-one in a neighborhood of 0 in g. Therefore π ◦exp is one-to- one in a neighborhood of 0 in g. Since exp is a local diﬀeomorphism, π is one-to-one in a neighborhood of e in G.

If dπ is an isomorphism, it is in particular a local diffeomorphism. Then for some sufficiently small −1 neighborhood U of e ∈ G, we have that π|U = exp ◦dπ ◦exp |U is a diffeomorphism of U with its image in H. Remark 3.31. The following three exercises are plagiarized from Knapp. They construct a smooth homomorphism π : SU(2) → SO(3) such that dπ : su(2) → so(3) is a Lie algebra isomorphism, but such that π is not invertible.

Here is the setup for the exercises: Observe that the subset S2 of R3 may be identiﬁed with the extended complex plane C ∪ {∞} via the following bijection:

x1 + ix2 (x1, x2, x3) 7→ z = . 1 − x3

Observe also that the group SU(2) acts on C ∪ {∞} by

αz + β α β w = g(z) = for g = ∈ SU(2). −βz + α −β α Exercise 16. Examine the preceding remark. 2 (1) Let (y1, y2, y3) denote the point in S which gets mapped to w via the bijection described. Show that

w + w w − w |w|2 − 1 y = , y = , and y = . 1 1 + |w|2 2 i(1 + |w|2) 3 |w|2 + 1

(2) Substitute in the result of (a) to show that SU(2) acts on S2 via the formula

   2 2 2 2    y1 Re(α − β ) −Im(α + β ) −2Reαβ x1 2 2 2 2  y2  =  Im(α − β ) Re(α + β ) −2Imαβ   x2  2 2 y2 2Reαβ −2Imαβ |α| − |β| x3

α β for g = ∈ SU(2). −β α 20 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Exercise 17. Interpreting the 3 × 3 matrix in the previous exercise as π(g), compute dπ on the su(2) i 0 0 1 0 i basis , , by diﬀerentiating π(exp tX) at t = 0 for each of these 2 × 2 0 −i −1 0 i 0 matrices X. Exercise 18. (1) Verify that dπ maps su(2) one-to-one onto so(3) in the previous exercise.

(2) Prove that π maps SU(2) onto SO(3) and is a diﬀeomorphism in a neighborhood of I.

(3) Verify that π has a nontrivial kernel. Remark 3.32. Let π : SU(2) → SO(3) be the map studied in the previous three exercises and let dπ be its diﬀerential. Since dπ is a linear isomorphism, it has an inverse map (dπ)−1 : so(3) → su(2). But since π is not invertible (due to its nontrivial kernel), there is no Lie group homomorphism ρ : SO(3) → SU(2) whose diﬀerential is (dπ)−1.

We will see later in the course that the only problem here is just that SO(3) is not simply connected. In fact whenever G is connected and simply connected and dπ : g → h is a Lie algebra homomorphism, then there exists a smooth homomorphism π : G → H whose diﬀerential is dπ. (See Theorem ...)

4. Automatic Smoothness 4.1. Preliminary Lemmas. Remark 4.1. If M and N are smooth manifolds and m ∈ M, n ∈ N, then there is a natural way to canonically identify the tangent space T(m,n)(M × N) with the product TmM × TnN. We leave it to the reader to check the details of this identiﬁcation, but we use it in the following lemma. Lemma 4.2. Let G be a Lie group, and let µ : G × G → G denote the multiplication map in G. Then the diﬀerential dµ(e,e) : TeG × TeG → TeG is precisely the addition mapping (Xe,Ye) 7→ Xe + Ye.

Proof. Since dµ is a linear mapping, it suﬃces to check that dµ(e,e)(Xe, 0) = Xe for any Xe ∈ TeG. To 0 see this, let c :(−, ) → G be a smooth curve in G with c(0) = e, c (0) = Xe. Deﬁne p = µ ◦ (c, e). Then by Lemma 3.24, for any f ∈ C∞(G), we have

0 dµ(e,e)(Xe, 0)(f) = (p (0))(f) d = f(µ(c(t), e) dt t=0 d = f ◦ c(t) dt t=0 = c0(0)(f),

whence dµe(Xe, 0) = Xe as claimed. Exercise 19. Let G be a Lie group and let ι : G → G denote the inversion map in G. Prove that the diﬀerential dιe : TeG → TeG is precisely the negation mapping Xe 7→ −Xe. Lemma 4.3. Let G be a Lie group and g its Lie algebra. If c(t) is a smooth curve in G with c(0) = e 0 and c (0) = Xe for X ∈ g, then there exists a δ > 0 so that

t k lim c = exp tX k→∞ k for all t ∈ (−δ, δ). LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 21

Proof. Repeat the proof of Lemma 2.16 nearly word for word, only replacing GL(n, C) with G and gl(n, C) with g, and canonically identifying g with some Euclidean space Rn. We did not use any- thing special about matrices in those arguments. The only substantive change should occur where we compute the value of Z0(0), because the calculus chain rule doesn’t make perfect sense in the context of general manifolds. To repair this part of the proof, simply use Corollary 3.26 and observe that 0 −1 0 Z (0) = d(exp )ec (0) = Xe. The rest of the arguments go through unchanged. Lemma 4.4. In any Lie group G, the mapping g 7→ g2, G → G is a local diffeomorphism at e. Proof. Observe that g 7→ g2 is the same mapping as g 7→ exp(2 exp−1 g), for g sufficiently close to e, because exp is a local diffeomorphism. Therefore g 7→ g2 is a composition of local diffeomorphisms, and thus a local diffeomorphism. 4.2. One-Parameter Subgroups. Definition 4.5. A one-parameter subgroup of a Lie group G is a continuous homomorphism φ : R → G. Corollary 4.6 (Corollary to Theorem 3.22). Let G be a Lie group and g its Lie algebra. For every X ∈ g, the mapping t 7→ exp tX (t ∈ R) defines a one-parameter subgroup of G. Proposition 4.7. Let G be a Lie group and g its Lie algebra. If φ is a one-parameter subgroup of G, then φ is the mapping t 7→ exp tX (t ∈ R) for some X ∈ g. In particular, φ is smooth. Proof. We know exp : g → G is a local diffeomorphism at e, and so is the mapping g 7→ g2 (G → G) by the previous lemma. So let V be an open neighborhood of e in G so small that exp is a diffeomorphism from a neighborhood of 0 in g onto V , and that g 7→ g2 is a diffeomorphism from a neighborhood of e in G onto V . Since φ is a homomorphism, φ(0) = e. Let U be an open neighborhood of 0 in g and V an open neighborhood of e in G so that exp : U → V is a diffeomorphism. By continuity of φ, there exists t0 ∈ R so small that φ(t0) ∈ U. Let Y ∈ g be such that φ(t0) = exp Y . Now since the squaring map is a bijection onto V , we get

1 φ(t0/2) = exp 2 Y

2 1 2 since φ(t0/2) = φ(t0) = exp Y = (exp 2 Y ) . Iterating this argument recursively, we obtain that

k 1 φ(t0/2 ) = exp 2k Y for every k ∈ N.

It follows then that, since φ is a homomorphism,

k k j 1 j j φ(jt0/2 ) = (φ(t0/2 )) = (exp 2k Y ) = exp( 2k Y ) for every j ∈ Z.

So we have shown φ(qt0) = exp qY for every dyadic rational q. Then by the continuity of φ and exp, 1 we have φ(tt0) = exp tY for all t ∈ . Setting X = Y , we have φ(t) = exp tX for all t ∈ . R t0 R Corollary 4.8. Let G be a Lie group. Then its Lie algebra g is in bijective correspondence with its space of one-parameter subgroups L(G). Theorem 4.9. Let G and H be Lie groups and let π : G → H be a continuous homomorphism. Then π is smooth. Proof. Since π is a continuous homomorphism, it carries one-parameter subgroups of G to one-parameter subgroups of H. Thus by Proposition 4.7, for every X ∈ g, there exists a unique element P (X) ∈ h such that 22 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

π(exp tX) = exp tP (X) for all t ∈ R.

We claim that P : g → h is actually a linear mapping. It is clear from the definition that P (λX) = λP (X) for any scalar λ ∈ R. To see that P is also additive, let X,Y ∈ g and consider the smooth curve c(t) = exp tX exp tY into G, which satisfies c(0) = e. Note that c = µ ◦ (p, q) where µ denotes multiplication in G, p(t) = exp tX and q(t) = exp tY . Computing the differential using Lemma 4.2, we get

0 0 0 c (0) = dµ(e,e)(p (0), q (0)) = Xe + Ye.

Then apply Lemma 4.3 and use the continuity of π to compute that for suﬃciently small t,

! t t k π(exp t(X + Y )) = π lim exp X · exp Y k→∞ k k t t k = lim π exp X π exp Y k→∞ k k t t k = lim exp P (X) · exp P (Y ) k→∞ k k = exp t(P (X) + P (Y )).

Therefore P (X + Y ) = P (X) + P (Y ), and so P : g → h is a linear mapping. In particular, P is a smooth mapping.

Now since exp is a local diﬀeomorphism, we see that near the identity in G, we have π = exp ◦P ◦exp−1, and therefore π is smooth near identity, i.e. there exists an open neighborhood V of e on which π|V is a smooth mapping.

Now let g ∈ G be arbitrary. Then on the open set gV containing g, we have π|gV = Lπ(g)◦π|V ◦Lg−1 |gV , so π|gV is a composition of smooth mappings and thus smooth. Since g was arbitrary, π is everywhere smooth. Remark 4.10. In this section we have shown that although we deﬁned one-parameter subgroups of Lie groups to be merely continuous, in fact they are automatically smooth. Moreover, any continuous homomorphism of Lie groups turns out to be automatically smooth. Then one may ask the question, do non-smooth (and hence non-continuous) homomorphisms of Lie groups exist? This goes back to an old question of Cauchy about whether the only homomorphisms from R to R+ are of the form t 7→ ekt.

To put the question into additional context, using Pettis’s theorem, one can prove that a homomorphism of Lie groups which is merely Borel measurable is automatically continuous, and hence automatically smooth. Similarly, Weil’s theorem implies that a homomorphism of Lie groups which is merely Haar measurable is automatically continuous, hence automatically smooth as well. So a non-smooth homomorphism must be very badly behaved indeed! In fact, it is consistent with the axioms of set theory (not including axiom of choice) that every subset of a Lie group is Haar measurable, which assumption would imply that all Lie group homomorphisms are smooth.

In the presence of the axiom of choice, however, there exist discontinuous Lie group homomorphisms. The easiest way to see this is to take R, view it as a vector space, and choose a Hamel basis for R over Q (this is a basis of cardinality continuum). Then a non-trivial permutation of this basis yields a vector space isomorphism from R to R, and in particular a group isomorphism, which is in general not LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 23

continuous (nor measurable in any sense). It follows that R embeds discontinuously into any Lie group of dimension ≥ 1.

On the other hand, some Lie groups do not admit any discontinuous group automorphisms! For example it is an old result of my dissertation advisor Kallman that any automorphism of the matrix a b group : a ∈ +, b ∈ is automatically continuous (hence automatically smooth). 0 1 R R

Corollary 4.11. Any topological group G may be made into a Lie group in at most one manner. More precisely, there is at most one smooth manifold structure on G under which the group operations are smooth.

Proof. Suppose G1 and G2 are two Lie groups such that id : G1 → G and idG2 → G are topological group isomorphisms. Then id : G1 → G2 is a topological group isomorphism, and in particular continuous, hence smooth. Similarly id : G2 → G1 is smooth, so G1 and G2 are isomorphic as Lie groups.

4.3. Cartan’s Theorem.

Lemma 4.12. Let G be a Lie group and g the Lie algebra of G, and let a, b be vector subspaces of g such that g = a ⊕ b. Then the mapping (A, B) 7→ exp A exp B, a ⊕ b → G is a local diﬀeomorphism at 0.

Proof. Let φ(A, B) = exp A exp B and compute the diﬀerential dφ. We have φ = µ ◦ (exp, exp), where µ denotes multiplication in G, and hence applying the chain rule and using Lemma 4.2 we have

dφ(0,0)(A, B) = dµ(e,e)(d exp0 A, d exp0 B) = A + B.

The diﬀerential dφ(0,0) is plainly invertible (because g = a ⊕ b), and therefore the inverse function theorem implies φ is a local diﬀeomorphism at the origin.

Theorem 4.13 (Cartan’s Theorem). Let G be a Lie group and H a closed subgroup of G. Denote h = {X ∈ g : exp tX ∈ H∀t ∈ R}. Then h is a vector subspace of g, and moreover there exists a neighborhood U of 0 in h and a neighborhood V of e in H such that exp is a homeomorphism of U onto −1 V . In particular, H may be regarded as a Lie group in which each pair (hV, exp ◦Lh−1 ) is a compatible chart, and dim H = dim h.

Proof. Find U an open neighborhood of 0 in g such that exp : U → exp(U) is a diﬀeomorphism, and in particular a homeomorphism. Note that V = U ∩ h is an open neighborhood of 0 in h, and exp V is a subset of H which contains e. We claim that exp V contains an open neighborhood of e in H.

Suppose it does not. Then we may find a sequence (hn) of elements which lie in H but not in exp(V ), so that hn → e. For sufficiently large n, we may write hn = exp Xn for Xn ∈ g. Now find a vector subspace s of g so that g = h ⊕ s. By Lemma 4.12, for sufficiently large n we may also find Yn ∈ h, Zn ∈ s such that exp Xn = exp Yn exp Zn, where Yn → 0 and Zn → 0. Since Xn ∈/ V , we have Zn 6= 0 for sufficiently large n.

Now identifying s with Euclidean space and equipping it with the Euclidean norm, note that the sequence (Zn/kZnk) lies in the unit sphere of s which is compact, and therefore it has a convergent subsequence. Passing to this subsequence, let us assume Zn/kZnk → Z for some Z ∈ s with kZk = 1. In particular, Z/∈ h.

−1 On the other hand since exp Xn ∈ H and exp Yn ∈ H, we have exp Zn = (exp Yn) exp Xn ∈ H. Fix kn any real number t, and let kn = bt/kZnkc. Since kZnk → 0, we have → 1. Therefore t/kZnk 24 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

kn lim (exp Zn) = lim exp knZn n→∞ n→∞ t = lim exp Zn n→∞ kZnk = exp tZ.

kn kn Since (exp Zn) = hn ∈ H for each n and H is topologically closed, we get exp tZ ∈ H. But this shows Z ∈ h after all, a contradiction!

Therefore exp V contains an open neighborhood of e in H. Restricting down to a smaller open subset of V , we see that exp : h → H is a local homeomorphism. Thus we have some chart (U, exp−1) at identity for H. Then, arguing as in Corollary 2.20, we can use the group operation in H to find charts −1 (hU, exp ◦Lh−1 ) at each h ∈ H, and these charts are smoothly compatible since G is a Lie group in the first place. Therefore H is a Lie group and dim H = dim h. Corollary 4.14. Let G be a Lie group and H a closed Lie subgroup of G, and let g and h denote their Lie algebras respectively. Then the space {X ∈ g : exp tX ∈ H∀t ∈ R} is canonically isomorphic with h, simply by restricting each vector field X on G to its subfield on H. Remark 4.15. Consequently we will generally identify h with a Lie subalgebra of g, as in the next corollary. Corollary 4.16. Let G be a Lie group with Lie algebra g, and let H and K be closed Lie subgroups of G. If H and K have the same Lie algebra (viewed as a subalgebra of g) then H0 = K0, i.e. the connected components of identity in H and K coincide.

Proof. If h is the Lie algebra of both H and K, then H0 = hexp hi = K0. Exercise 20. Let π : G → H be a smooth homomorphism of Lie groups, and let K = ker π. Show that K is a closed Lie subgroup of G, and that if k denotes the Lie algebra of K (viewed as a subalgebra of g the Lie algebra of G), then ker dπ = k.

5. Lie Subgroups and Lie Subalgebras 5.1. The Frobenius Theorems. Deﬁnition 5.1. Let M be an n-dimensional smooth manifold. For a natural number d ≤ n, a d-dimensional family D in TM is a mapping m 7→ Dm, where each Dm ⊆ TmM is a d-dimensional subspace.

A d-dimensional family is called smooth if for every m ∈ M, for every neighborhood U of m, there exist vector ﬁelds X1, ..., Xd on M such that for all u ∈ U, the vectors (Xi)u ∈ TuM span Du. We say these vector ﬁelds comprise a basis for D.

A vector ﬁeld X is called subordinate to D if Xm ∈ Dm for all m ∈ M. The family D is called involutive if whenever X and Y are subordinate to D, then the Lie bracket [X,Y ] is subordinate to D.

An integral manifold of the family D is a d-dimensional submanifold N such that, for each point n ∈ N, the tangent space TnN, identiﬁed with its image in TnM, is Dn. Remark 5.2. A natural way this situation described above can arise is as follows: let G be a Lie group and g the Lie algebra of G. Let h be a Lie subalgebra of g, i.e. a subspace which is closed under the Lie bracket. Then if we set Dg = {Xg : X ∈ h} for each g ∈ G, we have a smooth d-dimensional family in TM which is involutive. We would like to know, does there exist a Lie subgroup (in some suitable sense) H of G whose Lie algebra identiﬁes naturally with h? If so, H will be precisely an integral manifold of D. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 25

The answer turns out to be that we can’t get a full Lie subgroup H in general, but we can find a “local version’ via the local Frobenius theorem, which we prove below. More details on the applications later. Lemma 5.3. Let X be a vector field on a smooth n-dimensional manifold M, and let m ∈ M. Assume 1 n ∂ Xm 6= 0. Then there exist local coordinates (y , ..., y ) on a neighborhood V of m such that Xv = ∂yn (in the sense of Definition 3.1), for all v ∈ V . Sketch of Proof. Since this is a strictly local statement, it suffices to prove this in case M is an open subset of Rn and m = 0. In this case, we may write n X ∂ X = a · , i ∂x i=1 i

where each ai : M → R is a smooth function. Since X0 6= 0 by hypothesis, we have ai(0) 6= 0 for some i; assume without loss of generality that an(0) 6= 0.

n−1 Now for each ~u = (u1, ..., un−1) ∈ R , let us denote by c~u the integral curve for X with initial 1 n condition (u1, ..., un−1, 0), unique on its domain. So by deﬁnition, c~u = (c~u, ..., c~u) satisﬁes the equation

d i [ dt c~u(t)]t=t0 = ai(~u(t0)) for each 1 ≤ i ≤ n, for each t0 in the domain of c~u.

Deﬁne a map F : Rn → Rn by the rule

F (u1, ..., un−1, t) = c(u1,...,un−1)(t).

At this point in the proof sketch we appeal a bit to the general machinery of ordinary diﬀerential equations, as we did in the proof of Lie’s First Theorem 3.25. It is well-understood that the integral curves c~u vary smoothly with respect to the input parameter ~u, and from this information, one concludes that F is a smooth function itself.

Moreover if we write F = (F 1, ..., F n) in terms of its coordinate functions, and we restrict F to the line passing through ~u = (u1, ..., un−1) parallel to the xn-axis, then F is equal to c~u, which implies that

∂F i [ (t)]t=t = ai(F (u1, ..., un−1, t0) for each 1 ≤ i ≤ n, for each t0 in the domain of F . ∂xn 0

Now consider the partial derivatives of F = (F1, ..., Fn) evaluated at 0. First, note that F restricted to ∂Fi the hyperplane determined by xn = 0 is just the identity map by deﬁnition; and therefore (0) = ∂ij ∂xj ∂Fi for all 1 ≤ i ≤ n − 1, 1 ≤ j ≤ n. Secondly, by our remarks above, we have (0) = ai(0). So the xn Jacobian of F at 0 looks like

  1 0 ... 0 a1(0)  0 1 ... 0 a2(0)     ... .    0 0 ... 1 an−1(0)  0 0 ... 0 an(0)

This Jacobian has determinant an(0) 6= 0, and we conclude by the inverse function theorem that F is a diﬀeomorphism near 0. So let G = (y1, ..., yn): V → Rn denote the inverse map of F , deﬁned on some open neighborhood V of 0. By restricting if necessary, assume V ⊆ M. 26 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Now let f ∈ C∞(M) and v ∈ V , and compute using the chain rule:

∂ ∂ −1 n (f) = f ◦ G (x) ∂y ∂xn x=G(v) ∂ = f ◦ F (x) ∂xn x=G(v) n X ∂ ∂ = f(x) · F ∂x ∂x i i=1 i x=v n x=G(v) n X ∂ = f(x) · a (v) ∂x i i=1 i x=v

= Xv(f).

∂ Therefore X = ∂yn as claimed.

Lemma 5.4. If X1, ..., Xd are vector ﬁelds on a smooth manifold M such that [Xi,Xj] lies in the span of X1, ..., Xd for all 1 ≤ i, j ≤ d, and if for each m ∈ M we deﬁne Dm to be the span of (X1)m, .., (Xd)m, then D is an involutive family.

d X Proof. Any vector ﬁeld subordinate to D has the form fiXi in a neighborhood of a point m, where i=1 ∞ the fi’s are smooth functions deﬁned on this neighborhood. If X and Y are derivations of C M, verify the identity:

[fX, gY ] = fg[X,Y ] + fX(g)Y − gY (f)X for any smooth functions f and g,

which is obtained by applying both sides to an arbitrary h ∈ C∞(M) and performing a straightforward computation. Combining these two observations, we see that if X and Y are subordinate to D then so is [X,Y ]. Theorem 5.5 (Local Frobenius Theorem). Let D be a smooth involutive d-dimensional family in the tangent bundle of a smooth manifold M. Then for each point m ∈ M, there exists a neighborhood U of m and an embedded integral manifold N of D in U which contains m. If N 0 is another integral manifold of D containing m, then N and N 0 coincide in a neighborhood of m. Proof. Once again, since this is a purely local statement, it suﬃces for us to prove the theorem under the additional assumption that M is an open subset of Rn and m is the origin 0 ∈ Rn.

We proceed by induction on d. In case d = 1, the theorem is implies by the existence and local uniqueness of integral curves (Corollary 3.21). So assume d ≥ 2, and that the statement of the theorem is known to be true for all smaller-dimensional involutive families. Let X1, ..., Xd be a basis for D, and ∂ by Lemma 5.3, let us assume without loss of generality that Xd = . ∂xn n X ∂ ∂ For each 1 ≤ i < d, write Xi = hik for some smooth functions hik. Write Yi = Xi − hid ∂x , ∂xk n k=1 and set Yd = Xd. Then Y1, ..., Yd is still a basis for D, and

n−1 X ∂ Yi = hik for 1 ≤ i < d. ∂xk k=1 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 27

∂ Computing the Lie bracket of Yd = with Yi, and applying Clairaut’s theorem, we get ∂xn

n−1 ! n−1 ∂ X ∂ X ∂ ∂ [Yd,Yi] = ◦ hik − hik ◦ ∂xn ∂xk ∂xk ∂xn k=1 k=1 n−1 X ∂ ∂ ∂ ∂ ∂ ∂ = hik · + hik · ◦ − hik · ◦ ∂xn ∂xk ∂xn ∂xk ∂xk ∂xn k=1 n−1 X ∂hik ∂ = · . ∂xn ∂xk k=1

On the other hand since D is involutive, we may write the bracket [Yd,Yi] as a sum of products of ∂ smooth functions gij on M with the ﬁelds Y1, ..., Yd. By the above computation, the -component of ∂xn [Yd,Yi] is 0, so we may write

d−1 X [Yd,Yi] = gijYj. j=1

Next we show that if c1, ..., cd−1 are real constants, then there exist smooth functions f1, ..., fd−1 such that for small y1, ..., yn−1 we have fi(y1, ..., yn−1, 0) = ci, and

" d−1 # X Yd, fiYi = 0. i=1

To see that such functions exist, compute the bracket

" d−1 # d−1 d−1 d−1 X X ∂fi X X Y , f Y = Y + f g Y . d i i ∂x i i ij j i=1 i=1 n i=1 j=1

d−1 ∂fi X For this to be 0, we need the fi’s to be a solution to the ﬁrst-order system = − figij, which ∂xn i=1 exists locally with the prescribed initial condition.

d−1 X Now take (c1, c2, ..., cd−1) = (1, 0, ..., 0). Then the vector ﬁeld fiYi agrees with Y1 on the hyper- i=1 d−1 X plane xn = 0. Therefore we may replace Y1 with fiYi in our basis, and assume that [Yd,Y1] = 0. i=1 Repeating this process for each i, we may assume that [Yd,Yi] = 0 for all 1 ≤ i < d.

n−1 ∂h ∂ X ik ∂hik Recalling our computation above which showed [Yd,Yi] = · , we get that ∂x = 0, ∂xn ∂xk n k=1 i.e. the coefficients hik are independent of xn. Thus we can interpret Y1, ..., Yd−1 as defining d − 1 vector fields on Rn−1. By Lemma 5.4, they span a (d − 1)-dimensional smooth involutive family in n−1 R , and by the inductive hypothesis there exists an integral manifold N0 for this family. Then setting N = {(x1, ..., xn) ∈ M :(x1, ..., xn−1) ∈ N0, we obtain an integral manifold for D. 28 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Deﬁnition 5.6. Let D be a smooth d-dimensional family in the tangent bundle of a smooth manifold M, and let N be an integral manifold for D. N is called maximal if N is connected, and whenever N0 is a connected integral manifold for D which intersects N nontrivially, then N0 ⊆ N. Theorem 5.7 (Global Frobenius Theorem). Let D be a smooth involutive d-dimensional family in the tangent bundle of a smooth manifold M. Then for each point m ∈ M, there exists a unique maximal integral manifold N of D which contains m. Moreover the inclusion map id : N → M is an immersion, and N is second countable whenever M is.

Proof. If N0 and N1 are any two integral manifolds for D, then so is their intersection. Moreover, by the Local Theorem 5.5, every point in M is contained in some integral submanifold for D. Therefore the collection of all integral manifolds deﬁnes a base for some topology τ on M which is at least as ﬁne as the usual topology on M.

Let N denote the unique connected component of (M, τ) which contains m. By definition every point of N has a τ-open neighborhood diffeomorphic to Euclidean space where the charts are compatible on overlaps, so N is a smooth manifold. If n ∈ N is arbitrary and if U is any connected integral manifold of D in M containing n, then U is connected as a subspace of (M, τ), and hence U ⊆ N. Thus N is maximal. By construction, the differential of the inclusion map id : N → M is injective and hence N is an immersed submanifold of M.

It only remains to show that N is second countable if M is. If M is second countable, let {Uk : k ∈ N} be a countable basis for M. By the Local Theorem 5.5, for each n ∈ N, we may ﬁnd a τ-open neighborhood Nx of x in N such that Nx is an embedded submanifold of M and thus second countable, and such that Nx is a τ-connected component of N ∩ Uk for some k ∈ N. Let

Ck = {V ⊆ N : V is a τ-open second countable τ-connected component of N ∩ Uk},

[ and let C = Ck. Observe that C is a basis for (N, τ)(a priori, possibly an uncountable one). k∈N Let E ∈ C. We claim that there are only countably many E0 ∈ C for which E ∩ E0 6= ∅. For if not,

then by the pigeonhole principle, for some ﬁxed k0 ∈ N , we could ﬁnd uncountably many Eα ∈ Ck0 for which E ∩ Eα 6= ∅. But the Eα’s are pairwise disjoint τ-open sets since they are connected components

of Uk0 , and therefore the collection {E ∩ Eα} is an uncountable pairwise disjoint family of open sets in E, contradicting the fact that E is second countable.

0 Now pick F ∈ C arbitrarily. Define J0 = {F }, and recursively for each k ≥ 1, define Jk = {E ∈ C : 0 [[ E ∩ E 6= ∅ for some E ∈ Jk−1}. By our above remarks, each Jk is countable. Then B = Jk is a k∈N countable union of open second countable spaces, and hence open and second countable. If x ∈ B, then 0 since C is a basis for τ, there is E ∈ C so that x ∈ E and E ∩ B 6= ∅. Thus there is Jk and E ∈ Jk with 0 E ∩ E 6= ∅. Therefore E ∈ Jk+1, and by the definition of B, we get that E ⊆ B. This shows x ∈ B and hence B is closed. Since B is a clopen subset of N, we have N = B and N is second countable as claimed. 5.2. Subgroup-Subalgebra Correspondence. Definition 5.8. Let G be a Lie group. A Lie subgroup of G is a subgroup H which is a Lie group, and such that the inclusion mapping id : H → G is an immersion. Exercise 21. Let G be a group which is also a differentiable manifold, such that the group operations of multiplication and inversion are smooth. Show that the following are equivalent: (1) G is separable. (I.e. G is a Lie group.) (2) G is second countable. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 29

Theorem 5.9. Let G be a Lie group with Lie algebra g. Let h be a Lie subalgebra of g. Then there exists a unique connected Lie subgroup H of G such that h = {X ∈ g : exp tX ∈ H∀t ∈ R}. The subalgebra h is canonically isomorphic with the Lie algebra of H, via restricting the left-invariant vector ﬁelds in h to H. Moreover there exists an open neighborhood U of 0 in h and an open neighborhood V of e in H so that exp : U → V is a diﬀeomorphism.

Proof. Let X1, ..., Xd be a basis for h, and for each g ∈ G, let Dg be the subspace of TgG spanned by (X1)g, ..., (Xd)g. Since h is a Lie subalgebra it is closed under the Lie bracket, and hence D is a smooth involutive family in the tangent bundle of G. Thus by the global Frobenius Theorem 5.7, we may ﬁnd a unique maximal integral manifold H of D in G. We claim that H, with its group operations inherited from G, is a group.

To see this, observe that for any g, x ∈ G,

(dLg)g−1xDg−1x = span{(dLg)g−1x(X1)g−1x, ..., (dLg)g−1x(Xd)g−1x}

= span{(X1)x, ..., (Xd)x}

= Dx,

so D is left-invariant. It follows that if y, x ∈ H, then TxyH = TxLy(H) = (dLy)y−1xTy−1xH = (dLy)y−1xDy−1x = Dx, whence yH is an integral manifold for D. By maximality of H, yH ⊆ H, and therefore yx ∈ yH ⊆ H. This shows H is closed under multiplication. Similarly, x−1H is an integral manifold for D with e ∈ x−1H, which implies x−1 ∈ x−1H ⊆ H, and hence H is closed under inverses. So H is a subgroup of G. Since G is second countable, so is H, and H is an immersed submanifold of G. Therefore H is a Lie subgroup of G.

For any X ∈ g, the mapping t 7→ exp tX is the integral curve for X in G. Thus since H is the maximal integral manifold for h, it is clear that X ∈ h if and only if exp tX ∈ H for all t ∈ R, which proves the characterization of h given in the theorem. Moreover for a sufficiently small neighborhood W of 0 in g exp is a diffeomorphism, which sends U = W ∩ h onto an open d-dimensional submanifold of G which lies inside H. By the definition of the topology on H, this is an open neighborhood of e in H and we get that exp is a local diffeomorphism. 5.3. Local Groups and Local Homomorphisms. Definition 5.10. A local group is a topological space G together with an identity element e ∈ G, an open neighborhood U of e, and partially defined but continuous multiplication and inversion operations · : U × U → G and −1 : U → G which satisfy: (1) for all g, h, k ∈ G, if (gh)k and g(hk) are both defined then they are equal; (2) for all g ∈ U, g · e = e · g = g; and (3) for all g ∈ U, if g · g−1 is defined then it equals e, and if g−1 · g is defined then it equals e.

A local Lie group is a local group which is also a smooth manifold, whose partially deﬁned multiplication and inversion operations are both smooth. Remark 5.11. The easiest way to come by an example of a local group is to start with a topological group G, and restrict the group operations to a neighborhood of identity U in G. Perhaps surprisingly, not every local group arises in this manner; for instance see Tao’s Hilbert’s Fifth Problem Section 2.5.2. Corollary 5.12. Let G be a Lie group and g its Lie algebra. Consider two local subgroups of G to be equivalent if they coincide on a neighborhood of identity. To each local Lie subgroup H of G, we associate the tangent space TeH viewed as a subspace of TeG, and the Lie subalgebra h = {X ∈ g : Xe ∈ TeH}. Then any two equivalent local Lie subgroups H1 and H2 will deﬁne the same Lie subalgebra h, and every Lie subalgebra of g is associated to some local Lie subgroup H in this manner. In other words, this correspondence is a bijection between equivalence classes of local Lie subgroups H of G, and Lie subalgebras h of g. 30 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Definition 5.13. Let G and H be local groups. A local homomorphism from G to H is a mapping π : W → H, where W is an open neighborhood of identity in G on which the group operations are defined, which satisfies

(1) π(eg) = eh; (2) if x, y ∈ G and if xy is defined, then π(x)π(y) is defined in H and equals π(xy); and (3) if x ∈ G and if x−1 is defined, then π(x)−1 is defined in H and equals π(x−1).

We note that if π is a smooth local homomorphism of local Lie groups G and H, then we can still have enough data to construct the differential at identity dπe : TeG → TeH. Moreover if G and H are (global) Lie groups with corresponding Lie algebras g, h, and π : W → H is a smooth local homomorphism, we can still construct the differential dπ : g → h exactly as we have done previously, and the identity π ◦ exp = exp ◦dπ still holds in a neighborhood of 0 in g. Theorem 5.14 (Lie’s Second Theorem, First Version). Let G and H be Lie groups, and g and h the corresponding Lie algebras. Let φ : g → h be a Lie algebra homomorphism. Then there exists an open neighborhood W of identity in G and a smooth local homomorphism π : W → H such that dπ = φ. Proof. Consider the Lie group G × H, whose Lie algebra is g ⊕ h. Let k = {(X, φ(X)) : X ∈ g}, so k is the graph of φ. Since φ is a Lie algebra homomorphism, k is a Lie subalgebra of g ⊕ h, and therefore by Theorem 5.9 there exists a Lie subgroup K of G × H whose Lie algebra may be identified with k.

Let V ⊆ h be an open neighborhood of 0 so small that exp : V → exp(V ) is a diffeomorphism, and let U = φ−1(V ), so U is an open neighborhood of 0 in g. Without loss of generality assume U is small enough that exp : U → exp(U) is also a diffeomorphism. It follows immediately that (exp, exp) is a diffeomorphism of the neighborhood (U × V ) × k of 0 in k onto an open neighborhood of e in K.

Let Φg and Φh be the restrictions to k of the projection maps onto g and h, respectively. Since k is the graph of a linear mapping, Φg is a vector space isomorphism of k with g, and in particular a diﬀeomorphism. Likewise Φh is a linear mapping, hence smooth. Set W = exp(U), so W is an open −1 −1 −1 neighborhood of e in G. Deﬁne π = exp ◦Φh ◦ Φg ◦ exp = exp ◦φ ◦ exp on W . So π is a smooth mapping.

Now note that for g ∈ W , there is a unique Xg ∈ U such that exp(Xg) = g. It follows that for g ∈ W , −1 we have (g, h) ∈ K if and only if (g, h) = (exp Xg, exp ◦φ(Xg)), if and only if h = exp ◦φ◦exp (g) = π(g). In other words, the graph {(g, π(g)) : g ∈ W } of π is precisely an open neighborhood of e in K. There- fore the fact that π : W → K is a local homomorphism follows immediately from the fact that K is a subgroup of G × H.

Note also that locally near 0, by definition we have dπ = exp−1 ◦π ◦ exp = φ by Theorem 3.28. This is enough to conclude that dπ = φ since both maps are linear. Remark 5.15. We showed in the exercises that in general a mapping φ : g → h does not necessarily lift to a global group homomorphism π : G → H with dπ = φ, so the local result above is the best we can hope for. The place where this is reflected in the proof above is in paragraph 4: we need to guarantee that each g ∈ G corresponds to a unique h = π(g) ∈ H, and this may not be possible. It turns out that if G is a connected and simply connected group, then this problem may be repaired, which is our goal in the next section. Definition 5.16. Two Lie groups G and H are called locally isomorphic if there exist open neighborhoods U and V of identity in G and H, respectively, and a local homomorphism π : U → H which is a diffeomorphism of U onto V . Corollary 5.17. Any two Lie groups with isomorphic Lie algebras are locally isomorphic. Proof. If G and H are Lie groups and φ : g → h is an isomorphism of their Lie algebras, then there exist open neighborhoods of U and V of identity in G and H respectively, and smooth local homomorphisms LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 31

π : U → H and ψ : V → G such that dπ = φ and dψ = φ−1. By the chain rule for differentials, d(ψ ◦ π) = φ−1 ◦ φ = id. Hence, locally near e ∈ G, we have ψ ◦ π ◦ exp = exp ◦id = exp, and therefore ψ ◦ π = id locally near e (since exp is locally invertible). Similarly π ◦ ψ is the identity locally near e ∈ H. Since π and ψ are smooth and local homomorphic inverses of one another, we get that G and H are locally isomorphic as claimed. 6. Simply Connected Lie Groups 6.1. Discrete Subgroups and Quotients. Remark 6.1. In this section we will use the well-known fact that if G is a Hausdorff topological group and H is a closed normal subgroup of G, then G/H (viewed as both a group quotient and as a topological quotient space) is a Hausdorff topological group as well. Proposition 6.2. Let G be a separable topological group. (1) If H is an open subgroup of G then H is closed, and the quotient Q/H has the discrete topology. (2) The identity component G0 is open if G is locally connected. (3) If H is a discrete subgroup (i.e. discrete in its subspace topology) of G then H is closed in G. (4) If H is a closed discrete normal subgroup of G, and G is connected, then H lies in the center of G. Proof. (1) If H is open then the complement of H is a union of cosets of H, which are open. Hence H is closed. Since the canonical projection π : G → G/H is an open mapping, it follows that the singletons {xH} in G/H are open.

(2) In any locally connected space, the connected components are open.

(3) Choose U an open neighborhood of identity in G so that U ∩ H = {e}. Choose V an open 2 neighborhood of identity in G so small that V ⊆ U. Let x ∈ H, so there exists a sequence hn in H with −1 hn → x in G. By continuity of multiplication, hnx → e in G. Therefore there exists an N sufficiently −1 −1 −1 large that hN x ∈ V . Write hN x = v. Then by continuity of multiplication again, hN hn → v. −1 2 Hence we may find sufficiently large M such that for all n ≥ M, hN hn ∈ vV ⊆ V ⊆ U. But this −1 implies hN hn = e for all n ≥ M, i.e. hn = hN and the sequence is eventually constant. Therefore x = hN ∈ H, so H ⊆ H.

(4) Let h ∈ H. The mapping g 7→ ghg−1, G → H is continuous and therefore the image {ghg−1 : g ∈ G} is connected in H. But H is discrete, so this set is {e}, whence ghg−1 = e for all g ∈ G. Since h was arbitrary, H lies in the center of G. Deﬁnition 6.3. Let X and Y be connected separable metric spaces. Recall that a continuous onto map p : X → Y is called a covering map if for every y ∈ Y , there is an open neighborhood V of y such that each connected component of p−1(V ) is mapped homeomorphically via p onto V . Proposition 6.4. Let G be a connected, locally pathwise connected, separable topological group, and let H be a discrete subgroup of G. Then the natural quotient map π : G → G/H is a covering map. Proof. We already know π is open and surjective. Let U be a connected open neighborhood of e in G so small that U ∩ H = {e}. Let V be a connected open symmetric neighborhood of e in G such that 2 V ⊆ U. Form the open connected sets V h in G for h ∈ H. If h1, h2 ∈ H such that V h1 ∩ V h2 6= ∅, −1 −1 2 then there exist v1, v2 ∈ V such that v1h1 = v2h2, whence h1h2 = v1 v2 ∈ V ∩ H ⊆ U ∩ H = {e}, so h1 = h2. Therefore the V h’s are pairwise disjoint.

Next suppose g1, g2 ∈ V h for some ﬁxed h. If π(g1) = π(g2), then we have g1H = g2H and hence −1 −1 −1 −1 g1g2 ∈ H. But writing g1 = v1h, g2 = v2h for some v1, v2 ∈ V , we get that v1v2 = v1h · h v2 = −1 g1g2 ∈ H, whence v1 = v2 and g1 = g2. So π is one-to-one on V h, and therefore a homeomorphism of V h onto the open neighborhood VH of identity in G/H. Translating by each g ∈ G, we get that π is a homeomorphism of each open connected set gV h onto gV H in G/H, and hence π is a covering map as claimed. 32 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

6.2. Universal Covering Groups. Deﬁnition 6.5. If p : X → Y and p0 : X0 → Y are covering maps, and f : X → X0 is continuous such that p0 ◦ f = p, then f is called ﬁber-preserving.

Recall that a separable metric space X is called simply connected if given any closed loop in X (that is, a continuous path c : [0, 1] → M with c(0) = c(1)), c is homotopic to a trivial loop c0 which maps [0, 1] to a single point.

X is called locally simply connected if each x ∈ X has an open connected, simply connected neighborhood. Remark 6.6. We recall here without proof the basic theorems about covering spaces that we need. First recall that if f : X → Y is a continuous map of separable metric spaces and x ∈ X, then f induces a homomorphism f∗ : π1(X, x) → π1(Y, f(x)) by composing loops with f.

(1) (Map-Lifting Theorem) Let p : X → Y be a covering, and let x0 ∈ X, y0 ∈ Y such that p(x0) = y0. If Z is a connected, locally pathwise connected separable metric space, and if −1 g : Z → Y is continuous with z0 ∈ g (y0), then there exists a continuous f : Z → X with f(z0) = x0 and g = p◦f if and only if g∗(π1(Z, z0)) ⊆ p∗(π1(X, x0)). When f exists, it is unique.

(2) (Existence of Coverings) If Y is locally simply connected and y0 ∈ Y , and if H is a subgroup of π1(Y, y0), then there exists a covering space X with covering map p : X → Y and with point x0 such that p(x0) = y0 and p∗(π1(X, x0)) = H, where p∗ is one-to-one. If H = {e} then evidently X is simply connected. This space X is called the universal covering space of Y and is denoted Y˜ . This space is unique up to a ﬁber-preserving homeomorphism.

(3) (Manifold Structure on Coverings) If p : X → Y is a covering map and Y has the structure of a smooth manifold of dimension n, then X admits a unique smooth manifold structure in which dim X = n and the map p is smooth and everywhere regular. Moreover if Z is a smooth manifold and g : Z → Y is smooth, and f : Z → X is continuous with p ◦ f = g, then f is smooth. Theorem 6.7. Let G be a pathwise connected, locally connected, locally simply connected separable topological group, and let G˜ denote the universal covering space of G with covering map p : G˜ → G. Let e˜ ∈ p−1(e). Then there exists a unique multiplication ϕ˜ : G˜ × G˜ → G˜ on G˜ which makes G˜ into a separable topological group in such a way that e˜ is the identity element, and p is a group homomorphism. Proof. Let µ : G × G → G denote multiplication in G, and set ϕ = µ ◦ (p, p), ϕ : G˜ × G˜ → G. Since G˜ × G˜ is simply connected it has trivial fundamental group and hence by the map-lifting theorem, there is a unique continuous liftϕ ˜ : G˜ × G˜ → G˜ such that ϕ = p ◦ ϕ˜ andϕ ˜(˜e, e˜) =e ˜. We need to check that this deﬁnes a group operation on G˜.

To check associativity in G˜, ﬁrst use the associativity in G to compute that for all g, h, k ∈ G˜ we have

p ◦ ϕ˜(ϕ ˜(g, h), k) = ϕ(ϕ ˜(g, h), k) = ϕ(g, h) · p(k) = (p(g) · p(h)) · p(k) = p(g) · (p(h) · p(k)) = p ◦ ϕ˜(g, ϕ˜(h, k)).

Thusϕ ˜ ◦ (ϕ, ˜ id) andϕ ˜(id, ϕ˜) are lifts of the same map from G˜ × G˜ × G˜ → G, which each send (˜e, e,˜ e˜) to e ∈ G. By uniqueness of map lifts, they coincide. This impliesϕ ˜ is an associative multiplication.

Use a similar argument to check thate ˜ is an identity forϕ ˜: Consider the maps id : G˜ → G˜, ϕ˜ ◦ (id, e˜): G˜ → G˜, andϕ ˜ ◦ (˜e, id) : G˜ → G˜. Let F denote any of these three maps, and compute LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 33 that p ◦ F (˜e) = e and p ◦ F (g) = p(g) for every g ∈ G˜. Therefore uniqueness of map lifting implies all three maps are the same, and thuse ˜ is an identity element forϕ ˜ as claimed.

To ﬁnd inverse elements, let ˜ι be the unique lift of the mapping ι ◦ p : G˜ → G (where ι denotes inversion in G) which satisﬁes ˜ι(˜e) =e ˜. Thus for any g ∈ G˜ we have p(˜ι(g)) = p(g)−1. Therefore p ◦ ϕ˜ ◦ (id, ˜ι): G˜ → G is the constant function e on G˜. By uniqueness of map lifting,ϕ ˜ ◦ (id, ˜ι) coincides with the constant functione ˜ on G˜. In other words, ˜ι(g) is theϕ ˜-inverse of g for each g ∈ G˜.

Thus we have shown G˜ is a group, and p is a homomorphism by definition because p ◦ ϕ˜ = ϕ = µ ◦ (p, p). Exercise 22. Let G be a pathwise connected, locally connected, locally simply connected separable topological group. Suppose G˜1 and G˜2 are simply connected separable topological groups and p1 : G˜1 → G, p2 : G˜2 → G are covering maps which are also group homomorphisms. Prove that G˜1 is topologically isomorphic to G˜2. Definition 6.8. The above theorem and exercise imply that every pathwise connected, locally connected, locally simply connected separable topological group admits a simply connected separable topological group G˜ with a covering homomorphism p : G˜ → G, which is unique up to topological isomorphism. We call this group G˜ the universal covering group of G. Corollary 6.9. Let G be a Lie group. The universal covering group G˜ of G admits a unique smooth structure which makes it into a Lie group, and so that the covering homomorphism p : G˜ → G is smooth. Moreover the Lie algebras of G˜ and G coincide up to isomorphism, and dim G˜ = dim G. Proof. Let p : G˜ → G be the covering homomorphism. By our remark on manifold structures for coverings, G admits a dim G-dimensional smooth structure in which p is smooth and everywhere regular. Therefore the lift mapsϕ ˜ and ˜ι defined in the proof of Theorem 6.7 are smooth mappings, and hence G is a Lie group with this smooth structure. By Corollary 4.11, this Lie group structure is unique. Since p is everywhere regular, TeG˜ coincides with TeG and hence the groups have the same Lie algebra (up to isomorphism). Corollary 6.10. If G is a connected Lie group, then G is isomorphic to G/H˜ where G˜ is the universal covering group of G and H is some discrete central subgroup of G˜. Proof. Let p : G˜ → G be the covering homomorphism and set H = ker p. H is normal since it is the kernel of a homomorphism, discrete since p is a covering map, and central since it is discrete. 6.3. Extending Local Homomorphisms. Theorem 6.11 (Monodromy Principle). Let G and H be Lie groups, and assume G is connected and simply connected. Let U be an open neighborhood of identity in G. Then any smooth local homomorphism π : U → H can be extended to a smooth homomorphism π : G → H. Proof. Let dπ : g → h denote the differential of π and let k = {(X, dπ(X)) : X ∈ g} denote the graph of dπ in g × h. By Theorem 5.9, let K be the corresponding subgroup of G × H for which k is the Lie algebra of K.

By restricting U to a smaller open set if necessary, let us assume that exp−1 is deﬁned on U. Note that G = hUi since U is open, and thus every g ∈ G is of the form g = u1...un for u1, ..., un ∈ U. Con- −1 −1 sequently if X1 = exp u1, ..., Xn = exp un, then (g, π(u1)...π(un)) = (g, π(exp X1)...π(exp Xn)) = exp(X1, dπ(X1))... exp(Xn, dπ(Xn)) ∈ K. This argument shows that for every g ∈ G, there exists h ∈ H for which (g, h) ∈ K.

Let ΦG : K → G be restriction to K of the projection to the ﬁrst coordinate. Then ΦG is a smooth surjective group homomorphism. Also ker ΦG is a discrete normal subgroup of K since W ∩ker ΦG = {(e, e)}. This implies ΦG is a covering map by Proposition 6.4. But then since G is simply connected, this implies ΦG is a homeomorphism, and hence automatically a diﬀeomorphism since G and K are Lie groups. 34 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

−1 Let ΦH : K → H be restriction to K of the projection to the second coordinate. Then ΦH ◦ ΦG : G → H is the desired extension of π to all of G. Corollary 6.12 (Lie’s Second Theorem, Second Version). Let G and H be Lie groups, and g and h the corresponding Lie algebras. Let φ : g → h be a Lie algebra homomorphism. If G is connected and simply connected, then there exists smooth homomorphism π : W → H such that dπ = φ. Proof. Simply take the local homomorphism π guaranteed by Theorem 5.14 and extend it using Theorem 6.11. Corollary 6.13. Suppose G and H are connected, simply connected Lie groups which have isomorphic Lie algebras. Then G and H are isomorphic as Lie groups. Corollary 6.14. If G and G0 are connected Lie groups with isomorphic Lie algebras, then the universal covering groups G˜ and G˜0 are isomorphic as Lie groups.

7. Structure of Lie Algebras 7.1. Lie Algebra Representations. Remark 7.1. In this section we will use the definition of an abstract Lie algebra, which we gave long ago in Definition 2.9. Definition 7.2. Let g be a Lie algebra. An endomorphism of g is a linear mapping from g to itself. We denote the space of all endomorphisms of g by End(g). If g is finite-dimensional, then by choosing a basis of size n, one can represent each endomorphism as an n×n matrix in an obvious way. Therefore for finite dimensional g, we canonically identify End(g) with the matrix space gl(g) and treat these objects as interchangeable.

For any vector space V , End(V ) becomes a Lie algebra by deﬁning the bracket on End(V ) to be the commutator [X,Y ] = X ◦ Y − Y ◦ X.A Lie algebra representation of g on a vector space V is a Lie algebra homomorphism Φ : g → End(V ).

Given X ∈ g, we deﬁne the adjoint representation ad : g → End(g) of g on g by setting

(adX)Y = [X,Y ] for all X,Y ∈ g.

This mapping X 7→ adX is obviously linear by the bilinearity of the bracket, and it is a Lie algebra homomorphism because the bracket is antisymmetric and satisfies the Jacobi identity by definition. (See the below exercise.) Exercise 23. Given any Lie algebra g, prove that the adjoint representation ad : g → End(g) is a Lie algebra homomorphism. Remark 7.3. We offer the following theorem (Ado’s Theorem) as a hefty black box, so that we can deduce Lie’s Third Theorem as a corollary. The proof is algebraic and has a reputation for being fairly deep, so it is beyond our scope at the moment. I am not aware of any other tool which can be used to prove Lie’s Third Theorem. Theorem 7.4 (Ado’s Theorem). Every finite-dimensional Lie algebra g admits an injective representation on some finite-dimensional vector space V . Corollary 7.5 (Lie’s Third Theorem). Every finite-dimensional Lie algebra g is isomorphic to the Lie algebra of some Lie group G. Proof. Let ρ : g → gl(V ) be the representation guaranteed by Ado’s Theorem. Then ρ(g) is a Lie subalgebra of gl(V ) isomorphic to g, and gl(V ) is the Lie algebra of the Lie group GL(V ) of invertible linear transformations of V . By Theorem 5.9 there is a Lie subgroup G of GL(V ) whose Lie algebra is isomorphic to ρ(g), and hence to g. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 35

Corollary 7.6. Every finite-dimensional Lie algebra g is isomorphic to the Lie algebra of some simply connected Lie group G. Remark 7.7. At this point we have established the best possible correspondence for Lie groups and Lie algebras. Every Lie group determines a finite-dimensional Lie algebra and every finite-dimensional Lie algebra is determined by some Lie group (Corollary 7.5). The mapping from Lie groups to Lie algebras is not one-to-one; however if two Lie groups have the same Lie algebra then they are locally isomorphic (Corollary 5.17. Moreover this local isomorphism class of Lie groups contains a unique (up to isomorphism) connected simply connected Lie group (Corollary 6.9 and Corollary 6.13). Lastly, all connected Lie groups locally isomorphic to this connected simply connected Lie group are merely quotients of it by a discrete central subgroup (Corollary 6.10. Therefore, at this point we have determined that there are natural one-to-one correspondences between each of the following collections of objects: (1) Isomorphism classes of finite-dimensional Lie algebras; (2) Local isomorphism classes of Lie groups; (3) Isomorphism classes of connected simply connected Lie groups; (4) Classes of quotients of a particular connected simply connected Lie group by its discrete central subgroups. 7.2. Ideals. Definition 7.8. Let g be a Lie algebra and suppose a, b are subsets of g. We denote

[a, b] = span{[X,Y ]: X ∈ a,Y ∈ b}.

A Lie subalgebra h of g is called an ideal if [g, h] ⊆ h.

A Lie algebra g is called abelian if [g, g] = {0}. Example 7.9. Any vector space V may be made into an abelian Lie algebra by declaring that [X,Y ] = 0 for all X,Y ∈ V . Any two abelian Lie algebras of the same dimension are isomorphic as Lie algebras. Every one-dimensional Lie algebra is abelian, and therefore there is only one one-dimensional Lie algebra up to isomorphism. Exercise 24. Let g be a two-dimensional non-abelian Lie algebra. Show that there exists a basis {X,Y } for g under which [X,Y ] = Y . Deduce that up to isomorphism, there are only two distinct two-dimensional Lie algebras, an abelian one and a non-abelian one.

Example 7.10. The abelian two-dimensional Lie algebra is the Lie algebra of R2 and of the torus T2. a b The non-abelian two-dimensional Lie algebra is the Lie algebra of the ax+b group G = : a ∈ +, b ∈ . 0 1 R R Exercise 25. Classify, up to isomorphism, all connected two-dimensional Lie groups. Deﬁnition 7.11. If s is a subset of g then the space

Zg(s) = {X ∈ g :[X,Y ] = 0 for all Y ∈ s}

is called the centralizer of s in g. If s = {S} is a singleton, we may write Zg(S) instead of Zg(s). The centralizer of g is called the center of g and is denoted simply Z(g).

Similarly, the space

Ng(s) = {X ∈ g :[X,Y ] ∈ s for all Y ∈ s}

is called the normalizer of s in g.

Exercise 26. Let g be a Lie algebra and s a subset of g. Prove that Zg(s) is a subalgebra of g. Prove that if s is a subalgebra then Ng(s) is a subalgebra. Prove that if s is an ideal then Zg(s) is an ideal. 36 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Proposition 7.12. If a and b are ideals in a Lie algebra g, then so are a + b, a ∩ b, and [a, b]. Proof. The fact that a + b is an ideal follows immediately from the bilinearity of the Lie bracket. Showing that a ∩ b is an ideal is trivial. For [a, b], use the properties of the bracket to check that [X, [Y,Z]] = [[X,Y ],Z] + [Y, [X,Z]] for all X,Y,Z ∈ g, and therefore

[g, [a, b]] ⊆ [[g, a], b] + [a, [g, b]] ⊆ [a, b] + [a, b] ⊆ [a, b].

Exercise 27. Prove that if Φ : g → h is a Lie algebra homomorphism then ker Φ is an ideal in g. Definition 7.13. Let a be an ideal in a Lie algebra g. Then g/a as a vector space becomes a Lie algebra with the bracket defined by [X +a,Y +a] = [X,Y ]+a for all X,Y ∈ g. (Check this is well-defined because a is an ideal.) We call g/a the quotient algebra of g and a. The natural quotient map X 7→ X + a is a Lie algebra homomorphism, and therefore every ideal is the kernel of some Lie algebra homomorphism. Theorem 7.14 (Second Isomorphism Theorem for Lie Algebras). If g is a Lie algebra and a, b are ideals in g such that a + b = g, then g/a is isomorphic as a Lie algebra to b/(a ∩ b).

Proof. Define Φ : g/a → b/(a ∩ b) by the rule Φ(A + B + a) = B + (a ∩ b). If A1 + B1 = A2 + B2 then B1 −B2 = A1 −A2 ∈ a∩b, whence B1 +a∩b = B2 +a∩b, so the map Φ is well-defined. It is clearly linear. If B1 +a∩b = B2 +a∩b, then B1 = A+B2 for some A ∈ a∩b, so B1 +a = A+B2 +a and Φ is one-to-one. Φ is clearly onto as well. To check that Φ is a Lie algebra homomorphism, use the definition of the Lie bracket on g/a and the fact that a is an ideal to compute that [A1 + B1 + a,A2 + B2 + a] = [B1,B2] + a. Then the fact that Φ preserves the bracket is obvious. Definition 7.15. Let g be a finite-dimensional Lie algebra. Define g0 = g, and recursively for each j ≥ 0, define gj+1 = [gj, gj]. Then the decreasing sequence g0 ⊇ g1 ⊇ g2 ⊇ ... is called the commutator series for g. By Proposition 7.12, each term in the sequence is an ideal in g. We say that g is solvable if gj = {0} for some integer j.

Similarly, define g0 = g and recursively set gj+1 = [g, gj] for each natural number j. This defines a decreasing sequence g0 ⊇ g1 ⊇ g2 ⊇ ... called the lower central series for g. Each gj is an ideal in g. We say that g is nilpotent if gj = 0 for some integer j. Example 7.16 (Standard Examples of Solvable and Nilpotent Lie Algebras). (1) The space of all upper triangular real (or complex) matrices is a solvable Lie algebra. (2) The space of all upper triangular matrices with 0’s along the diagonal is a nilpotent Lie algebra. Exercise 28. Show that any subalgebra or Lie-homomorphic image of a solvable Lie algebra is solvable. Similarly, any subalgebra or Lie-homomorphic image of a nilpotent Lie algebra is nilpotent. Proposition 7.17. If a is a solvable ideal in a Lie algebra g, and g/a is solvable, then g is solvable. Proof. Let π : g → g/a be the quotient map, and assume (g/a)k is trivial. Since π is a Lie algebra homomorphism, π(gj) = (g/a)j for each j. Thus π(gk) is trivial, whence gk ⊆ a. But a` = {0} for some k+` k ` ` `, whence g = (g ) ⊆ a = {0}, and g is solvable. Proposition 7.18. If g is a finite-dimensional Lie algebra, then there exists a unique solvable ideal r of g which contains all solvable ideals of g. Proof. Since g is finite-dimensional it suffices to check that any sum h = a + b of solvable ideals (which is an ideal by Proposition 7.12) is solvable. Since a is a solvable ideal in h, by the Second Isomorphism Theorem 7.14 we have h/a ≡ b/(a ∩ b). The latter quotient is solvable since b is solvable. Therefore h/a is solvable. Since a and h/a are solvable, h is solvable by Proposition 7.17. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 37

Deﬁnition 7.19. The ideal r given in the previous proposition is called the (solvable) radical of g, and is denoted rad g.

A finite-dimensional Lie algebra g is called simple if g is nonabelian and has no proper nonzero ideals. The Lie algebra g is called semisimple if g has no nonzero solvable ideals, i.e. if rad g = {0}. Proposition 7.20. If g is a simple Lie algebra then [g, g] = g. Every simple Lie algebra is semisimple. Every semisimple Lie algebra g has Z(g) = {0}. Proof. If g is simple, then since [g, g] is an ideal, it is either {0} or g. It can’t be the former because g is nonabelian. Similarly, rad g is either {0} or g. If the latter holds, then g is solvable, whence [g, g] is a proper subset of g, which is impossible. So rad g = {0} and g is semisimple. If g is semisimple, then Z(g) is an abelian ideal and hence a solvable ideal, which proves Z(g) = {0}. Proposition 7.21. If g is a finite-dimensional Lie algebra then g/rad g is semisimple. Proof. Assume h is a solvable ideal in g/rad g. Let π : g → g/rad g be the quotient mapping and consider the ideal a = π−1h in g. Then a/(a ∩ rad g) ≡ h is solvable, and a ∩ rad g is solvable, whence a is solvable by Proposition 7.17. Therefore a ⊆ rad g and h = π(a) is trivial. This proves g/rad g is semisimple. Exercise 29. Show that every two-dimensional Lie algebra is solvable. Exercise 30. Show that every three-dimensional Lie algebra is either simple or solvable. Exercise 31. Show that so(3) and su(2) are simple Lie algebras. 7.3. Connections with the Structure of Lie Groups. Definition 7.22. Let g be a Lie algebra. We denote by Aut(g) the group of all vector space automorphisms of g. If g is finite-dimensional, then we may choose a finite basis for g, and we naturally identify Aut(g) with the matrix group GL(g) of invertible transformations of g with respect to the chosen basis. Note that in this case GL(g) is a Lie group, and its Lie algebra is isomorphic to the space gl(g) of endomorphisms of g. Moreover the exponential map exp : gl(g) → GL(g) is the matrix exponential.

Now let G be a Lie group and g its Lie algebra. For each g ∈ G, let cg : G → G denote conjugation −1 by g, i.e. cg(x) = gxg for x ∈ G. It is plain that cg is a smooth isomorphism from G to G. So we can deﬁne its diﬀerential, which we denote by Ad(g) = dcg. Thus Ad(g): g → g is a Lie algebra isomorphism.

Moreover, if g, h ∈ G then cgh = cg ◦ ch, and it follows from the chain rule for diﬀerentials that Ad(gh) = Ad(g)Ad(h). Therefore the mapping Ad : G → GL(g) is a group homomorphism. Observe −1 also that for a ﬁxed X ∈ g, in a neighborhood of e ∈ G, we have Ad(g)X = exp ◦cg ◦ exp (X), whence g 7→ Ad(g)X is smooth near e ∈ G. Applying this observation to elements X of a basis for g, we get that Ad : G → GL(g) is smooth near e ∈ G. But this is enough to imply that Ad : G → GL(g) is smooth −1 everywhere, since Ad(g) = LAd(g) ◦Ad◦Lg at each g ∈ G. Therefore Ad is a Lie group homomorphism, called the adjoint representation of G on g. Proposition 7.23 (Taylor’s Theorem on a Lie Group). Let G be a Lie group and g its Lie algebra. Let ∞ c(t) = exp tX for t ∈ R. Then for all f ∈ C (G), for all g ∈ G, for all t0 ∈ R, we have

n dn (X f)(g · c(t0)) = n (f(g exp tX)) . dt t=t0

Moreover, if k · k is a norm on g (for instance the Euclidean norm) then for X suﬃciently close to 0 we have

n X 1 f(exp X) = (Xkf)(e) + R (X), k! n k=0

n+1 where kRn(X)k ≤ CnkXk for some constant Cn. 38 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

∞ 0 Proof. Since c is the integral curve for X, for f ∈ C (G), we have (Xf)(c(t0)) = Xc(t0)f = c (t0)(f) = d f(c(t)) , which proves the ﬁrst part of the proposition for n = 1 and g = e. Replacing f with dt t=t0 f ◦ Lg proves it for general g. Then replacing f with Xf and iterating the argument proves it for n ≥ 1.

For the second part of the proposition, write out f ◦ c in Taylor series about t = 0 and evaluate at t = 1 to get:

f(exp X) = f ◦ c(1) n n+1 X 1 dk 1 Z 1 d = f(exp tX) + (1 − s)n f(exp sX)ds k! dtk n! ds k=0 t=0 0 n X 1 1 Z 1 = (Xkf)(e) + (1 − s)n(Xn+1f)(exp sX)ds. k! n! k=0 0

X For the remainder term on the right above, write X = λiXi in terms of a basis for g, and expand Xn+1. Since X is restricted to lie in a compact set, so is exp sX, and so the integral on the right which n+1 depends on X is dominated by | max{λi}| times a ﬁxed integral which does not depend on X. This proves the estimate in the proposition. Lemma 7.24. Let G be a Lie group with Lie algebra g. If X,Y ∈ g then

1 2 3 (1) exp tX exp tY = exp[t(X + Y ) + 2 t [X,Y ] + O(t )], and (2) exp tX exp tY (exp tX)−1 = exp[tY + t2[X,Y ] + O(t3)]

as t → 0. Here O(t3) denotes a smooth function on t suﬃciently small which is bounded and remains bounded when divided by t3. Proof. Since exp is invertible near 0, write exp tX exp tY = exp Z(t) for some smooth function Z, for 2 3 t small. Since Z(0) = 0, we have Z(t) = tZ1 + t Z2 + O(t ), and we wish to compute Z1 and Z2. If f ∈ C∞(G) is arbitrary, then by Taylor’s Theorem (Proposition 7.23) we have

2 X 1 f(exp Z(t)) = [(tZ + t2Z + O(t3))kf](e)+)(t3) k! 1 2 k=0 1 = f(e) + t(Z f)(e) + t2 Z2 + Z f (e) + O(t3). 1 2 1 2

On the other hand, another application of Taylor’s Theorem gives

2 X 1 f(exp tX exp sY ) = sk(Y kf)(exp tX) + O (s3) k! t k=0 2 2 X X 1 1 = skt`(X`Y kf)(e) + O (s3) + O(t3), k! `! t k=0 `=0

3 3 3 where Ot(s ) is a smooth function which depends on t, but for which the bound on O(s )/s may be taken independent of t for t suﬃciently small. Setting s = t, we get

2 1 2 1 2 3 f(exp Z(t)) = f(exp tX exp tY ) = f(e) + t[(X + Y )f](e) + t [( 2 X + XY + 2 Y )f](e) + O(t ). LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 39

Now replacing f with f ◦ Lg for each g ∈ G, we deduce the equality of operators Z1 = X + Y and 1 2 1 2 1 2 1 2 Z1 + Z2 = 2 X + XY + 2 Y . From here compute that Z2 = 2 [X,Y ], which proves statement (1) of the theorem. Statement (2) follows from applying (1) twice.

Theorem 7.25. Let G be a Lie group and g its Lie algebra, and let Ad : G → GL(g) and ad : g → gl(g) denote the adjoint representations of G and g respectively. Then dAd = ad, and therefore Ad ◦ exp = exp ◦ad.

0 Proof. Fix X,Y ∈ g and denote c(t) = exp tX, so c (0) = Xe. By Lemma 3.24, to compute dAd it suﬃces to compute the derivative of the smooth curve Ad ◦ c in GL(g) at 0.

By Lemma 7.24, for t ∈ R we have Ad(c(t))tY = Ad(exp tX)tY = tY + t2[X,Y ] + O(t3). Dividing by t, we have

Ad(c(t))Y = Ad(exp tX)Y = Y + t[X,Y ] + O(t2).

Diﬀerentiating at t = 0, we get dAd(X)Y = [X,Y ] = ad(X)Y as claimed.

Proposition 7.26. Let G be a Lie group and g its Lie algebra. If X,Y ∈ g with [X,Y ] = 0, then exp X exp Y = exp Y exp X = exp(X + Y ).

Proof. We have that ad(X)Y = 0, and thus an easy induction shows that adn(X)Y = 0 for each positive integer n ≥ 1. Compute then that

Ad(exp X)Y = exp(adX)Y ∞ X 1 = adk(X)Y k! k=0 = Y.

Then since Ad(exp X) is the diﬀerential of cexp X , we get that

−1 exp X exp Y (exp X) = cexp X (exp Y ) = exp(Ad(exp X)Y ) = exp(Y ).

This implies that exp X exp Y = exp Y exp X. More generally, this argument implies that exp tX exp uY = exp uY exp tX for every t, u ∈ R.

Now consider the smooth map q(t, u) = exp tX exp uY , q : R2 → G. Since c(u) = exp uY , c : R → G ∞ is the integral curve for the left-invariant vector ﬁeld Y ∈ g, for any t0, u0 ∈ R and for any f ∈ C (G), we have 40 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

∂ d f(q(t, u)) = (f ◦ L )(c(u)) ∂u du exp t0X (t,u)=(t0,u0) u=u0 0 = c (u0)(f ◦ Lexp t0X )

= Yc(u0)(f ◦ Lexp t0X )

= (dLexp t0X )c(u0)Yc(u0)(f)

= Yexp t0X·c(u0)(f)

= Yexp t0X exp u0Y (f).

Similarly, since we may also write q(t, u) = exp uY exp tX, we can compute that for any t0, u0 ∈ R, for any f ∈ C∞(G), we have

∂ f(q(t, u)) = X (f). ∂t exp t0X exp u0Y (t,u)=(t0,u0)

∞ Now by the chain rule, we have for any v0 ∈ R, for any f ∈ C (G),

d ∂ ∂ f(q(v, v)) = f(q(t, u)) + f(q(t, u)) dv ∂u ∂t v=v0 (t,u)=(v0,v0) (t,u)=(v0,v0)

= Yexp v0X exp v0Y (f) + Xexp v0X exp v0Y (f).

Consequently, v 7→ q(v, v) is exactly the integral curve for X + Y with initial condition 0 7→ e. There- fore q(v, v) = exp v(X+Y ) by uniqueness of integral curves, and setting v = 1 proves the proposition. Theorem 7.27 (First Isomorphism Theorem for Lie Algebras). If φ : g → h is a homomorphism of Lie algebras and k = ker φ, then g/k is isomorphic as a Lie algebra to h.

Proof. Verify that X + k 7→ φ(X) is the desired isomorphism. Proposition 7.28. Let G be a Lie group and g its Lie algebra. (1) If a Lie subgroup H of G is normal then h = {X ∈ g : exp tX ∈ H∀t ∈ R} is an ideal in g. Conversely if H is a Lie subgroup of G and h is an ideal, then the connected component H0 is normal in the connected component G0. (2) If G is abelian then g is abelian. If g is abelian then G0 is abelian. (3) If G is simple then g is simple. If g is simple then any normal Lie subgroup of G0 is discrete and central.

Proof. (1) Assume H is normal in G and let g ∈ G, Y ∈ h. Since exp tY ∈ H for each t ∈ R, we have

exp(tAd(g)Y ) = exp(Ad(g)(tY ))

= exp(dcg(tY ))

= cg(exp tY ) ∈ H,

and therefore Ad(g)Y ∈ h. This shows Ad(g)h ⊆ h for each g ∈ G. Consequently, for a ﬁxed X ∈ g, the map t 7→ Ad(exp tX)Y is a smooth curve in g. Now applying Lemma 3.24 write LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 41

[X,Y ] = ad(X)Y d = Ad(exp tX) Y dt t=0 1 = lim [Ad(exp hX)Y − Y ]. h→0 h

The right-hand side is a limit of terms involving only vector space operations on h. Since h is a vector subspace of g, it is closed topologically, and therefore this limit lies in h. Consequently, [X,Y ] ∈ h for all X ∈ g, and h is an ideal as claimed.

Conversely, assume h is an ideal. Then an easy induction shows that ad(X)nY ∈ h for all X ∈ g, Y ∈ h. Consequently

∞ X 1 ad(X)nY ∈ h, k! k=0

because h is topologically closed. But this expression just says exp(ad(X))Y ∈ h, and therefore Ad(exp X)Y ∈ h. Since G0 is generated by exp g and since Ad is a homomorphism, we get that Ad(g)Y ∈ h, for all g ∈ G0, for all Y ∈ h. This implies that cg(exp Y ) = exp(Ad(g)Y ) ∈ H for every g ∈ G0, for every Y ∈ h. Lastly, since H0 is generated by exp h and since cg is a homomorphism, we conclude that cg(h) ∈ H for every g ∈ G0, h ∈ H0. This shows H0 is normal in G0.

(2) If G is abelian then Ad(g): g → g is the identity for every g ∈ G, whence Ad : G → GL(g) is the trivial homomorphism. It follows immediately that ad : g → gl(g) is the trivial Lie algebra homomorphism, and g is abelian. Conversely if g is abelian then [X,Y ] = 0 for all X,Y ∈ g, and consequently exp X exp Y = exp Y exp X for all X,Y ∈ g by Proposition 7.26. Since G0 = hexp gi, it follows that G0 is abelian.

−1 (3) Assume G is simple. Observe that for any g ∈ G, the subgroup gG0g of G is connected and clopen and hence coincides with G0. In other words G0 is normal in G, and since G is simple we have that G = G0 is a connected group.

Let h be a proper ideal in g, and let H be the connected Lie subgroup of G whose Lie algebra is h. Then H is normal in G by part (1). Therefore H = {e} since H is proper. So h = {0} and g is simple.

Conversely if g is simple and H is a proper normal Lie subgroup of the connected component G0, then h = {X ∈ g : exp tX ∈ H∀t ∈ R} is a proper ideal in g and hence h = {0}. Thus H0 = {e} and H is discrete in G0, hence central in G0.

Deﬁnition 7.29. A connected Lie group G is called a simple Lie group if it has no non-trivial connected normal subgroups. Note that by the previous theorem, if the Lie algebra g of a Lie group G is simple, then G0 is a simple Lie group. Conversely, it is easy to see that if G is a simple Lie group then g is simple. In particular, if G is connected then G is a simple Lie group if and only if g is simple.

Example 7.30. SU(2) is an example of a simple Lie group which is not a simple group, because {I, −I} is a closed normal discrete central subgroup of SU(2).

Corollary 7.31 (Classiﬁcation of Connected Abelian Lie Groups). (1) Every connected abelian Lie group is isomorphic to a group of the form R` × (R/Z)k for some `, k ∈ N. (2) Every compact connected abelian Lie group is isomorphic to a torus Tk = (R/Z)k. 42 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Proof. Every connected abelian Lie group of dimension n has Lie algebra Rn, and is therefore a quotient k n n X R /D by some discrete subgroup D of R . This group has the form D = Zvi where the vi’s are i=1 k ! n n X n−k linearly independent vectors in R . Writing R = Rvi ⊕ R and quotienting by D, we obtain i=1 the first conclusion. The second conclusion follows because if the group is compact, we must have ` = 0. 7.4. Semidirect Products. Definition 7.32. Let a and b be Lie algebras and let g = a ⊕ b be the direct sum of a and b as vector spaces. We can define a bracket on g which agrees with the brackets on a and b when restricted to these subalgebras respectively, and which gives [A, B] = 0 for all A ∈ a, B ∈ b. We say that g is the Lie algebra direct sum of a and b. Here a, b are ideals in g.

A derivation of a Lie algebra g is an endomorphism D ∈ End(g) which satisﬁes the rule D[X,Y ] = [X,DY ] + [DX,Y ] for all X,Y ∈ g.

We denote the space of all derivations of g by Der(g). It is straightforward to check that Der(g) is a Lie subalgebra of End(g). It is also straightforward to check that ad(X) ∈ Der(g) for all X ∈ g, and thus ad : g → Der(g) is a Lie algebra homomorphism. Proposition 7.33. Let a and b be Lie algebras and suppose π : a → Der(b) is a Lie algebra homomorphism. Then there exists a unique Lie algebra structure on the vector space direct sum g = a ⊕ b whose bracket operation agrees with those on the subalgebras a and b, and satisfying [A, B] = π(A)(B) for all A ∈ a, B ∈ b. Furthermore this structure makes b into an ideal of g. Proof. The deﬁnition of the bracket on g is already given in the proposition, so we need only check the Jacobi identity. Let X,Y,Z ∈ g. If all three are in either a or b, then we are done. By skew-symmetry, we reduce to two cases:

Case 1: Suppose X,Y ∈ a and Z ∈ b. Then π([X,Y ]) = π(X)π(Y ) − π(Y )π(X) since π is a Lie algebra homomorphism. Applying both sides of this equality to Z, we get [[X,Y ],Z] = [X, [Y,Z]]+[Y, [X,Z]] which is equivalent to the Jacobi identity.

Case 2: Suppose X ∈ a and Y,Z ∈ b. Since π(X) is a derivation of b, we have π(X)[Y,Z] = [π(X)Y,Z] + [Y, π(X)Z]. Therefore [X, [Y,Z]] = [[X,Y ],Z] + [Y, [X,Z]], which is equivalent to the Ja- cobi identity.

It is clear also that b is an ideal under this bracket. Deﬁnition 7.34. Let a, b, π, and g be as in Proposition 7.33. We call g the semidirect product of a and b (with respect to π) and we write g = a ⊕π b. Example 7.35 (Examples of Semidirect Products). (1) If a, b are any two Lie algebras and π : a → Der(b) is the trivial homomorphism π(a) = 0, then a ⊕π b is just the direct sum a ⊕ b. (2) Let g denote the nonabelian two-dimensional Lie algebra, with basis {X,Y } satisfying [X,Y ] = Y . Then letting a be the span of X and b be the space of Y , we have g = a ⊕ad b. Deﬁnition 7.36. Let G, H be Lie groups with Lie algebras g, h respectively. A mapping τ : G×H → H is a smooth action of G on H if the mapping is smooth, and if g 7→ τ(g, ·) is a homomorphism of G into the automorphism group Aut(H) of H. In this case the usual semidirect product group G nτ H becomes a Lie group with G × H as its underlying manifold.

For a smooth action τ : G × H → H, we denote by τ(g) the diﬀerential of τ(g, ·): H → H. So τ(g): h → h, and τ : G → GL(h). Via arguments completely analogous to those we gave for Ad in LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 43

Deﬁnition 7.22, we see that τ : G → GL(h) is a smooth homomorphism. Its diﬀerential dτ is a Lie algebra homomorphism from g into Der(g), which allows us to form the Lie algebra semidirect product g ⊕dτ h. Theorem 7.37. Let G, H be Lie groups with g, h as Lie algebras respectively, and let τ : G × H → H be a smooth action. Then the Lie algebra of G nπ H is g ⊕dτ h.

Proof. The tangent space at (e, e) in G nτ H is g ⊕ h as a vector space, and the inclusions of G and H exhibit the bracket structure on g and h as corresponding to the respective bracket structures on g ⊕ {0} and {0} ⊕ h. We have to check the brackets of members of g ⊕ {0} with members of {0} ⊕ h. Let X ∈ g, Y ∈ h, and write X˜ = (X, 0) and Y˜ = (0,Y ). Then

exp(Ad(exp tX˜)sY˜ ) = (exp tX˜)(exp sY˜ )(exp tX˜)−1 = (exp tX, e)(e, exp sY )(exp(−tX), e) = (exp tX, e)(exp(−tX), exp sY ) = (e, τ(exp tX) exp sY ).

For ﬁxed t, both sides above are one-parameter subgroups, and the corresponding identity on the Lie algebra level is Ad(exp tX˜)Y˜ = (0, τ(exp tX)Y ).

Diﬀerentiating both sides at t = 0 yields [X,˜ Y˜ ] = (0, dτ(X)Y ).

Theorem 7.38. Let G and H be connected simply connected Lie groups with Lie algebras g and h respectively. Let π : g → Der(h) be a Lie algebra homomorphism. Then there exists a unique smooth action τ : G × H → H such that dτ = π, and G nτ H is a connected simply connected Lie group with Lie algebra g ⊕π h. Proof. Since G is connected and simply connected, by Lie’s Second Theorem we can find a smooth τ : G → GL(h) such that dτ = π. For fixed g the map τ(g): h → h is a Lie algebra automorphism. Therefore since H is connected and simply connected, by Lie’s Second Theorem again we can find a smooth group automorphism τ(g): H → H with d(τ(g)) = τ(g).

If g1, g2 ∈ G then d(τ(g1g2)) = τ(g1g2) = τ(g1)τ(g2) = d(τ(g1)τ(g2)). So τ(g1g2) and τ(g1)τ(g2) have the same diﬀerential, whence they coincide. This shows τ : G → Aut(H) is a group homomorphism.

It remains only to check that τ : G × H → H is smooth. First observe that since τ : G → GL(h) is smooth, it follows immediately that the mapping (g, Y ) 7→ τ(g)Y , G × h → h is smooth. Choose an open neighborhood W of e in H so that exp is a diﬀeomorphism of some neighborhood of 0 in h onto W . Then τ is smooth on G × W , because τ is just the composition (g, exp Y ) 7→ (g, Y ) 7→ τ(g)Y 7→ exp(τ(g)Y ) = τ(g, exp Y ).

For h ∈ H, deﬁne τ h : G → H by τ h(g) = τ(g, h) for g ∈ G. To see that τ h is smooth, observe that H = hW i since H is connected and W is open, and therefore we may write h = h1h2...hn for h h1 hn h1, ..., hn ∈ W . Since h 7→ τ(g, h) is an automorphism of H, we have τ (g) = τ (g)...τ (g). Since each τ hi is smooth, we get that τ h is smooth.

Finally observe that for any ﬁxed h, τ restricted to G × W h is the composition (g, wh) 7→ (g, w) 7→ (τ(g, w), τ h(g)) 7→ τ(g, w)τ(g, h) = τ(g, wh) 44 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

which is smooth. Since G × H is covered by open sets of the form G × W h, h ∈ H, this shows that τ is smooth everywhere on G × H. 7.5. Solvable Lie Algebras. Proposition 7.39. An n-dimensional Lie algebra g is solvable if and only if there exists a ﬁnite sequence of subalgebras g = a0 ⊇ a1 ⊇ ... ⊇ an = {0} such that for each i, ai+1 is an ideal in ai and dim(ai/ai+1) = 1. j Proof. Assume g is solvable. Form the commutator series g and interpolate vector subspaces ai in the j j+1 sequence so that dim(ai/ai+1) = 1 for all i. Thus for any i, we may ﬁnd j such that g ⊇ ai ⊇ ai+1 ⊇ g . Then j j j+1 [ai, ai] ⊆ [g , g ] = g ⊆ ai+1, and therefore ai is a subalgebra for each i, and ai+1 is an ideal in ai.

0 Conversely, let the sequence exist. Choose Xj such that aj = RXj + aj+1. We have g = a0. Assume j j+1 j j inductively that g ⊆ aj, and consider j + 1. We have g = [g , g ] ⊆ [RXj + aj+1, RXj + aj+1] ⊆ j [RXj, aj+1] + [aj+1, aj+1] ⊆ aj+1. Therefore g ⊆ aj for each j, and the solvability of g follows from an = {0}.

Remark 7.40. The sequence g = a0 ⊇ a1 ⊇ ... ⊇ an = {0} whose existence was shown in Proposition 7.39 is called an elementary sequence for the solvable Lie algebra g. Note that for each i, since ai+1 is an ideal of codimension 1 in ai, we may write ai = RXi ⊕ad ai+1, which displays ai as a semidirect product of a one-dimensional subalgebra with ai+1. Consequently, the solvable Lie algebras are exactly those which can be built up by starting with a 1-dimensional abelian Lie algebra and building up, one dimension at a time, with semidirect products. Corollary 7.41. If g is a solvable Lie algebra of dimension n, then there exists a connected simply n X connected Lie group G with Lie algebra g, and G is diﬀeomorphic to g via the mapping tiXi 7→ i=1 n Y exp tiXi, for a suitable choice of basis {X1, ..., Xn} for g. Moreover, there exists a sequence of closed i=1 connected simply connected subgroups G = G0 ⊇ G1 ⊇ ... ⊇ Gk = {e} such that each Gi is a semidirect product Gi = R nτi Gi+1 with Gi+1 normal in Gi. Remark 7.42. In particular, the previous corollary shows that exp is onto for connected simply connected Lie groups G. If G is not simply connected, this is not necessarily true even for solvable G.

Proof. By Proposition 7.39, we may find an elementary sequence of Lie algebras g = a0 ⊇ a1 ⊇ ... ⊇ an = {0}. For each i we have that ai+1 is an ideal of codimension 1 in ai, which means we may write ai as a semidirect product of RXi with ai+1 (where Xi is a member of ai not in ai+1). Using R as a simply connected Lie group whose Lie algebra is isomorphic to RXi, we invoke Theorem 7.38 to recursively define each Gi as a semidirect product of R with Gi+1 (starting with Gn = {e}). Each group Gi is then diffeomorphic to ai via the mapping in the statement of the corollary. Corollary 7.43. A connected simply connected solvable Lie group is diffeomorphic to Rn. Corollary 7.44. A connected solvable Lie group is diffeomorphic to R` × Tk for some k, ` ≥ 0, and a compact connected solvable group is diffeomorphic to Tk for some k ≥ 0. Corollary 7.45. If G is a Lie group with a solvable Lie algebra g, then G is solvable.

Proof. Because G is a quotient of a semidirect product of the form given in Corollary 7.41. Theorem 7.46 (Lie’s Theorem). Let g be a solvable Lie algebra. Let V 6= {0} be a ﬁnite-dimensional

vector space over C, and denote by EndC(V ) the Lie algebra of all C-linear maps from V to V , with the standard Lie bracket. Let π : g → EndC(V ) be a Lie algebra homomorphism. Then there exists a vector v ∈ V which is a simultaneous eigenvector for all the members of π(g). LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 45

Proof. By induction on dim g. If dim g = 1 then π(g) consists of all multiples of a single endomorphism, and so the theorem is true for the base case.

By induction, assume the result holds for all solvable Lie algebras of dimension strictly less than g. Since g is solvable, we have [g, g] 6= g; choose a subspace h of codimension 1 in g so that [g, g] ⊆ h. Then [g, h] ⊆ [g, g] ⊆ h and h is an ideal. Also h is solvable. By the inductive hypothesis, let w ∈ V be a nonzero simultaneous eigenvector for π(h). It means that π(X)w = λ(X)w for each X ∈ h, where λ : g → C is a scalar-valued function.

p Fix Y ∈ g such that g = RY + h. Set w−1 = 0; w1 = w; and wp = π(Y ) w for each p ≥ 1. Let W = span{wi : i ≥ 0}. Then π(Y )W ⊆ W . Let v be an eigenvector for π(Y ) in W . We claim that v is an eigenvector for each π(X), X ∈ h.

In order to show this, ﬁrst set Wp = span{w0, ..., wp−1}. We will show that for every X ∈ h, when we look at the quotient space V/Wp, we have

π(X)wp + Wp = λ(X)wp + Wp.

We show this by induction on p. It holds for p = 0 by deﬁnition of wp = w0. Then assume it for p, and consider p + 1. Applying the inductive hypothesis in the third line below, we have

π(X)wp+1 + Wp+1 = π(X)π(Y )wp + Wp+1

= π([X,Y ])wp + π(Y )π(X)wp + Wp+1

= λ([X,Y ])wp + π(Y )λ(X)wp + Wp+1

= λ(X)π(Y )wp + Wp+1

= λ(X)wp+1 + Wp+1.

This completes the induction on p.

The argument above implies that π(X)W ⊆ W for each X ∈ h. In particular, for the element [X,Y ] ∈ h, by choosing a basis for W from among {w0, w1, ...}, we may assign π([X,Y ]) a matrix representation in which each diagonal entry is λ([X,Y ]). Thus tr W π([X,Y ]) = λ([X,Y ]) dim W (where the trace is computed for the matrix acting on W ). But tr W π([X,Y ]) = tr W ([π(X), π(Y )]) = 0, and therefore λ([X,Y ]) = 0 after all.

Next we reﬁne our previous argument to show that in fact, π(X)wp = λ(X)wp for each X ∈ h, for each p. This holds for p = 0 by deﬁnition. Assume it for p, and consider p + 1:

π(X)wp+1 = π(X)π(Y )wp

= π([X,Y ])wp + π(Y )π(X)wp

= π(Y )λ(X)wp

= λ(X)π(Y )wp

= λ(X)wp+1.

This shows π(X)wp = λ(X)wp for all p as claimed. Therefore each wp is a simultaneous eigenvector for π(h), and hence W consists entirely of simultaneous eigenvectors for π(h). In particular, v is a simultaneous eigenvector for each π(X) ∈ π(h), and for π(Y ) as well. Since g = RY + h, we have proven the theorem. 46 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Corollary 7.47. Let g be a solvable Lie algebra. Let V 6= {0} be a ﬁnite-dimensional vector space over

C, and let π : g → EndC(V ) be a Lie algebra homomorphism. Then there exists a sequence of subspaces V = V0 ⊇ V1 ⊇ ... ⊇ Vm = {0} such that π(g)Vi ⊆ Vi for each i, and dim(Vi/Vi+1) = 1. Consequently V has a basis with respect to which all the matrices of π(g) are upper triangular. α β Exercise 32. For each A = ∈ GL(n, ), let g be a three-dimensional vector space with a γ δ R A basis {X,Y,Z}, and define a bracket on gA by [X,Y ] = 0; [X,Z] = αX + βY ; and [Y,Z] = γX + δY .  0 0 1   0 0 0  (1) Show that gA is a Lie algebra, by showing that X ↔  0 0 0 ; Y ↔  0 0 1 ; Z ↔ 0 0 0 0 0 0  α γ 0   β δ 0  gives an isomorphism with a Lie algebra of matrices. 0 0 0 (2) Show that gA is solvable but not nilpotent. (3) Take δ = 1 and β = γ = 0. Show that for different choices of α > 1, the Lie algebras gA are pairwise non-isomorphic. Deduce that there are uncountably many non-isomorphic solvable three-dimensional Lie algebras. Definition 7.48 (Heisenberg Lie Algebra). The Heisenberg Lie algebra is the Lie algebra of matrices     0 a b  defined by g =  0 0 c  : a, b, c ∈ R .  0 0 0  Exercise 33. Let g be a three-dimensional Lie algebra. Prove that if dim[g, g] = 1, then either g is isomorphic to the Heisenberg Lie algebra, or else g is isomorphic to the direct sum of R with the nonabelian two-dimensional Lie algebra.  0 0 0  Exercise 34. Compute the Lie bracket of so(3) with respect to the basis ~i =  0 0 1 ; ~j = 0 −1 0  0 0 1   0 1 0   0 0 0 ; ~k =  −1 0 0 . −1 0 0 0 0 0

1 0 Exercise 35. Compute the Lie bracket of sl(2, ) with respect to the basis X = ; Y = R 0 −1 0 1 0 0 ; Z = . 0 0 1 0 7.6. Nilpotent Lie Algebras. Theorem 7.49 (Engel’s Theorem). Suppose g is Lie subalgebra of gl(V ) for some ﬁnite-dimensional vector space V 6= {0}, and suppose every element of g is nilpotent (in the sense of Exercise 11. Then g is a nilpotent Lie algebra. Moreover, there exists v ∈ V such that Xv = 0 for all X ∈ g, and therefore there is a basis for V in which all X ∈ g are upper triangular with 0’s on the diagonal. Proof. By induction on dim g. If dim g = 1 then all the conclusions are trivial. So let dim g > 1, and assume the theorem holds for all smaller-dimensional Lie subalgebras. We will prove that there exists v ∈ V with Xv = 0 for all X ∈ g. From here, the existence of a basis for which all X are upper triangular with 0’s on the diagonal is the standard argument from linear algebra, and then the fact that g is a nilpotent Lie algebra is immediate because a subalgebra of a nilpotent Lie algebra is nilpotent.

Let h be a proper subalgebra of g of maximal dimension. By hypothesis, h is nilpotent. Deﬁne a representation ρ : h → gl(g/h) by the rule

ρ(H)(X + h) = ad(H)X + h = [H,X] + h for all X + h ∈ g/h. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 47

We claim that each ρ(H) is nilpotent. To see this, note that for H ∈ h, the operator ad(H) is the diﬀerence L(H) − R(H) of the commuting operations of left and right multiplication by H. Therefore the binomial theorem implies that

2m X 2m (adH)2mX = (−1)j L(H)jR(H)2m−jX j j=0 2m X 2m = (−1)j HjXH2m−j. j j=0

If Hm = 0 then every term on the right-hand side above is 0, whence adH is nilpotent. It follows immediately that ρ(H) is nilpotent as claimed.

Since dim ρ(h) < dim g, by the inductive hypothesis there exists a coset X0 + h 6= h such that ρ(H)(X0 + h) = h for all H ∈ h. Put otherwise, we have [H,X0] ∈ h for all H ∈ h. But then, the vector space RX0 + h is closed under the bracket and is hence a subalgebra. Since h was chosen to have maximal dimension, we must have g = RX0 + h and therefore h has codimension 1 in g. Moreover, we have shown h is an ideal.

Let V0 = {v ∈ V : Hv = 0∀H ∈ h}. By induction, V0 6= {0}. Note that for any H ∈ h, for any v ∈ V0, we have HX0v = [H,X0]v + X0Hv = 0 + X0 · 0 = 0. Therefore X0V ⊆ V . By assumption X0 is nilpotent, and 0 is its only eigenvalue. So there exists an eigenvector v0 ∈ V0 with X0v0 = 0. This v0 is the simultaneous eigenvector for all of g with gv0 = 0. Lemma 7.50. If g is a Lie algebra, then g is nilpotent if and only if adg is nilpotent. Proof. If g is nilpotent, then adg is nilpotent because ad is a Lie algebra homomorphism. Conversely, suppose adg is nilpotent, and let (adg)k = {0}. If X1, ..., Xk,Xk+1 ∈ g then we have

[[[...[Xk+1,Xk], ..., X3],X2],X1] = ad[[...[Xk+1,Xk], ..., X3],X2]X1

= [[...[adXk+1, adXk], ..., adX3], adX2]X1

= 0 · X1 = 0.

Therefore gk+1 = {0} and g is nilpotent. Corollary 7.51. A Lie algebra g is nilpotent if and only if ad(X) is a nilpotent operator for every X ∈ g. Proof. The forward direction is clear. For the reverse direction, if ad(X) is always nilpotent then ad(g) is a nilpotent Lie algebra by Engel’s Theorem 7.49, and hence g is nilpotent by Lemma 7.50. Theorem 7.52. If g is a solvable Lie algebra, then [g, g] is nilpotent.

Proof. By Ado’s theorem, we may assume g is a Lie subalgebra of gl(n, R). Then by Lie’s Theorem 7.46 g is conjugate in gl(n, C) to a Lie algebra h of upper triangular matrices in gl(n, C). Compute directly that [h, h] consists of upper triangular matrices with 0’s along the diagonal, which is an algebra of nilpotent operators. Therefore [g, g] is nilpotent by Engel’s Theorem 7.49. Theorem 7.53. If g is a solvable Lie algebra, then g has a unique maximal nilpotent ideal n, called the nilradical of g. Proof. Set n = {X ∈ g : adX is nilpotent}. By Corollary 7.51, n is nilpotent, and by Corollary 7.51 again, n is maximal among nilpotent subalgebras. To see that n is an ideal, note that [g, n] ⊆ [g, g] ⊆ n by Theorem 7.52. (Thanks to C. Martin for the simpliﬁed proof.) 48 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Exercise 36. Let g be a three-dimensional Lie algebra. Prove that if dim[g, g] = 2, then g is isomorphic to gA for some A ∈ GL(2, R) as in Exercise 32. (Hint: Use Theorem 7.52 to show that [g, g] is abelian. Let {X,Y } be a basis for [g, g] and extend to a basis {X,Y,Z} for g. Deﬁne α, β, γ, δ by [X,Z] = αX + βY α β and [Y,Z] = γX + δY . Show that is invertible.) Conclude that the only nilpotent three- γ δ dimensional Lie algebras are the abelian one and the Heisenberg Lie algebra, and the only other solvable three-dimensional Lie algebras are those given by Exercise 32. Exercise 37. Let g be a three-dimensional simple Lie algebra.

(1) Using Engel’s theorem, choose X0 ∈ g such that adX0 is not nilpotent. Show that the one- dimensional subspace RX0 has a complementary subspace invariant under adX0. (2) Show by linear algebra that some real multiple X of X0 is a member of a basis {X,Y,Z} where either (i) (adX)Y = 2Y and (adX)Z = −2Z; or else (ii) (adX)Y = −Z and (adX)Z = Y . (3) Writing [Y,Z] in terms of the basis and applying the Jacobi identity, show that Y can be multiplied by a constant so that the ﬁrst case leads to an isomorphism with sl(2, R) and the second case leads to an isomorphism with so(3). Deduce that the only simple three-dimensional Lie algebras are sl(2, R) and so(3). Theorem 7.54 (Baker-Campbell-Dynkin-Hausdorﬀ Formula). Let G be a Lie group and g its Lie algebra. There exists an open neighborhood U of 0 in g such that for X,Y ∈ U the series

X X (−1)k (adX)p1 (adY )q1 ...(adX)pk (adY )qk (adX)m X ∗ Y = X + · Y (k + 1)(q1 + ... + qk + 1) p1!q1!...pk!qk!m! k,m≥0 pi+qi>0 converges, and satisfies exp X exp Y = exp(X ∗ Y ). Remark 7.55. The formula in the previous theorem is named after Baker, Campbell, and Hausdorff who proved around the turn of the century that the operation ∗ : g×g → g, X ∗Y = exp−1(exp X exp Y ) can be expressed abstractly as a convergent infinite series using only commutators, commutators of commutators, etc. These authors did not provide an explicit formula. The formula written above with numerical coefficients was first given by Dynkin in 1947.

Here are the ﬁrst few terms of the BCDH series written out:

1 1 1 X ∗ Y = X + Y + [X,Y ] + ([X, [X,Y ]] + [Y, [Y,X]]) − [Y, [X, [X,Y ] 2 12 24 1 − ([Y, [Y, [Y, [Y,X]]]] + [X, [X, [X, [X,Y ]]]]) 720 1 + ([X, [Y, [Y, [Y,X]]]] + [Y, [X, [X, [X,Y ]]]]) 360 1 + ([Y, [X, [Y, [X,Y ]]]] + [X, [Y, [X, [Y,X]]]]) + ... 120

Observe that if X and Y commute i.e. [X,Y ] = 0 then the BCDH formula implies exp X exp Y = exp(X + Y ) a la our Proposition 7.26. Also replacing X,Y with tX, tY the BCDH implies formula (1) given in Lemma 7.24.

The proof of the BCDH formula would require too long a diversion for this semester-long course, so we refer the interested reader to Hilgert and Neeb’s Structure and Geometry of Lie Groups Section 9.2.5. Corollary 7.56. Let g be a finite-dimensional nilpotent Lie algebra. Then the Baker-Campbell-Dynkin- Hausdorff operation ∗ : g × g → g is a polynomial map which defines a smooth group operation on g. Thus (g, ∗) becomes a connected simply connected nilpotent Lie group with Lie algebra isomorphic to g, and with exp : g → (g, ∗) equal to the identity map. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 49

Proof. It is clear from the deﬁnition that ∗ is a polynomial map because g is nilpotent and thus ∗ is smooth. By inspection, X ∗ 0 = 0 ∗ X = X and X ∗ (−X) = (−X) ∗ X = 0 for all X ∈ g.

Now let G be a Lie group whose Lie algebra is g. Then there exists a neighborhood U of 0 in g such that exp X exp Y = exp(X ∗ Y ) for all X,Y ∈ U, and for which exp is a diﬀeomorphism on U. By the continuity of ∗, choose an open neighborhood V of 0 such that (V ∗ V ) ∗ V and V ∗ (V ∗ V ) are both subsets of U. Then for X,Y,Z ∈ V we have exp((X ∗ Y ) ∗ Z) = exp X exp Y exp Z = exp(X ∗ (Y ∗ Z)) by the associativity of the group operation on G. Since exp is invertible on V , we conclude that (X ∗ Y ) ∗ Z = X ∗ (Y ∗ Z), whenever X,Y,Z ∈ V . But both maps (X,Y,Z) 7→ (X ∗ Y ) ∗ Z and (X,Y,Z) 7→ X ∗ (Y ∗ Z) are polynomial maps, and since they agree on an open set, they agree everywhere. This shows ∗ is associative and thus (g, ∗) becomes a Lie group with identity 0 and inverses given by X−1 = −X.

In addition, it is clear from the definitions that exp : (g, ∗) → G is a local Lie group isomorphism. So (g, ∗) and G have the same Lie algebra, i.e. the Lie algebra of (g, ∗) is isomorphic to g. Corollary 7.57. Let G be a connected Lie group and g its Lie algebra. If g is nilpotent, then the exponential map exp : (g, ∗) → G is the universal covering homomorphism. In particular, exp is surjective. Proof. As in the proof of the previous corollary, observe that exp : U → G is a local Lie group isomorphism, where U is a sufficiently small neighborhood of 0 in (g, ∗). Then by the Monodromy Principle Theorem 6.11, exp extends to a global covering homomorphism π :(g, ∗) → G. But since each one- dimensional subspace of g is also a one-parameter subgroup of (g, ∗), it is easy to see that π must coincide with exp. Proposition 7.58. If g is a nilpotent Lie algebra, then the Lie group (g, ∗) is nilpotent. Proof. To simplify notation, let us denote G = g viewed as a Lie group with group multiplication given by ∗. We have that exp : g → G is a diffeomorphism with exp(0) = e ∈ G. For each of finitely many integers i, let gi denote the i-th term in the lower central series for g. Let Hi = exp gi = gi. Since gi is a subspace of the finite-dimensional space g, it is topologically closed and therefore Hi is closed in G. Since gi is an ideal in g it follows from the definition of ∗ that gi is closed under ∗, and therefore Hi is a closed subgroup of G. We claim that if Gi denotes the i-th term in the lower central series for the group G, then Gi ⊆ Hi and thus the nilpotency of G follows from the nilpotency of g.

We show this by induction on i, assuming that Gi−1 ⊆ Hi−1 for i ≥ 1. Let X ∈ G, Y ∈ Gi−1. Then Y ∈ Hi−1 = gi−1 by induction. Check by the deﬁnition of ∗ that the group commutator [X,Y ] = X ∗ Y ∗ (−X) ∗ (−Y ) is purely a ﬁnite sum of terms involving brackets of X with Y . Consequently X ∗ Y ∗ (−X) ∗ (−Y ) ∈ [g, gi−1] = gi = Hi, whence Gi ⊆ Hi. This completes the induction and the proof. Corollary 7.59. If G is a connected Lie group with nilpotent Lie algebra g, then G is nilpotent.

Proof. Because G is a quotient of (g, ∗). Proposition 7.60. The (Lie-algebraic) center of a nilpotent Lie algebra g coincides with the (group- theoretic) center of (g, ∗). Proof. Let z be the center of g and Z the center of (g, ∗). The inclusion z ⊆ Z is immediate from the deﬁnition of ∗. Conversely if X ∈ Z, then id = Ad(exp X) = exp(adX), whence adX = 0 since exp (viewed as a mapping from gl(g) to GL(g) is injective by Exercise 11. Therefore X ∈ z. Corollary 7.61. If G is a connected Lie group with nilpotent Lie algebra g, then the center of G is connected. Proof. Because exp : (g, ∗) → G is the universal covering homomorphism and the center of (g, ∗) is a subspace of g, hence connected. Corollary 7.62. Every compact connected Lie group G with nilpotent Lie algebra g is isomorphic to a torus Tk. 50 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Proof. G is isomorphic to (g, ∗)/Γ where Γ is a discrete subgroup of the group-theoretic center Z of (g, ∗). But Z is also the Lie-algebraic center of g, and by the definition one sees that ∗ is the same thing as + when restricted to Z. So Γ is a discrete additive subgroup of g. Since G is diffeomorphic to g/Γ and the former is compact, we must have that span(Γ) = (g). But Γ ⊆ Z, so g ⊆ Z and g is an abelian Lie algebra. Therefore G is a torus. 7.7. The Killing Form. Definition 7.63. Let g be a finite-dimensional Lie algebra. The Killing form on g is the symmetric bilinear mapping B : g × g → R given by B(X,Y ) = tr (adXadY ), for X,Y ∈ g.

The Killing form is invariant in the sense that B((adX)Y,Z) = −B(Y, (adX)Z) for all X,Y,Z ∈ g. Example 7.64 (Computing the Killing Form). Let g be the non-abelian two-dimensional Lie algebra. So g has a basis {X,Y } with [X,Y ] = Y . To understand the Killing form, it suffices to compute its value on the basis vectors. So need to compute the traces of ad(X)ad(X), ad(X)ad(Y ), and ad(Y )ad(Y ). Fixing the basis {X,Y }, the matrix representation of ad(X)ad(X) is 0 0 ad(X)ad(X) = 0 1 and therefore B(X,X) = 1. Similarly compute that B(X,Y ) = 0, B(Y,Y ) = 0. Therefore in general we have B(aX + bY, cX + dY ) = ac. Definition 7.65. Let B denote the Killing form on a finite-dimensional Lie algebra g. The radical of B is rad B = {X ∈ g : B(X,Y ) = 0 for all Y ∈ g}.

One can also view rad B as the intersection of the kernels of the mappings X 7→ B(X,Y ), for Y ∈ g. B is called nondegenerate if rad B = {0}. Proposition 7.66. In any finite-dimensional Lie algebra g with Killing form B, rad B is an ideal in g. Proof. It is clear from the bilinearity of B that rad B is a subspace. The fact that rad B is an ideal follows from the identity B([X,Y ],Z) = B((adX)Y,Z) = −B(Y, (adX)Z) = B(Y, [X,Z]). Exercise 38. Compute the Killing form of sl(2, R) and show that it is nondegenerate. Exercise 39. Show that if B is the Killing form of sl(2, R), then B(X,X) is a multiple of det X independent of the choice of X. Exercise 40. Consider so(3) and fix the basis of Exercise 34. Viewing elements of so(3) as column  a  vectors a~i + b~j + c~k =  b , show that the Killing form on so(3) is a multiple of the dot product. c Theorem 7.67 (Cartan’s Criterion for Solvability). A finite-dimensional Lie algebra g is solvable if and only if [g, g] ⊆ rad B, where B denotes the Killing form for g. Proof. In order to simplify the notation in this proof, we adopt the following convention: instead of regarding ad as a map ad : g → gl(g), we view it as a map ad : g → gl(n, C) for some nonnegative integer n. This may be accomplished without loss of generality, by choosing a suitable isomorphism of gl(g) with gl(n, R), and identifying it with its image in gl(n, C). This adoption does not change the Killing form since the trace is invariant under a change of basis.

For the forward direction, assume g is solvable. By Lie’s Theorem 7.46, there exists a basis of Cn in which all matrix representations of elements of ad(g) are upper triangular. Then members of ad[g, g] = [ad(g), ad(g)] have 0’s on the diagonal, and so for X ∈ g and Y ∈ [g, g], adXadY has 0’s on the diagonal. Therefore B(X,Y ) = 0, and [g, g] ⊆ rad B as claimed. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 51

Conversely, assume [g, g] ⊆ rad B. We will show that ad[g, g] consists of nilpotent operators, and therefore [g, g] is nilpotent by Engel’s Theorem 7.49. Consequently g is solvable.

Let Y ∈ [g, g] and assume for the sake of contradiction that ad(Y ) is not nilpotent. Since ad(Y ) lies in gl(n, C), we may apply the following sharp form of the Jordan decomposition for complex matrices: there exist unique s, n ∈ gl(n, C) such that ad(Y ) = s + n, where s is diagonalizable, n is nilpotent, and sn = ns. Moreover s = p(ad(Y )) for some polynomial p without constant term.

By our assumption that ad(Y ) is not nilpotent, we have s 6= 0. Let λ1, ..., λm be the distinct eigen- values of s, and let V1, ..., Vm be the corresponding eigenspaces. Deﬁne s ∈ gl(n, C) to be λi on Vi. Then m X 2 tr (ss) = |λi| > 0. i=1 Since s and n commute, so do s and n, and therefore sn is nilpotent. It follows that tr (sad(Y )) = tr (ss) + tr (sn) = tr (ss) > 0.

u X Next, since Y ∈ [g, g], write Y = [Xi,Zi] with Xi,Zi ∈ g. Then i=1 u X tr (sad(Y )) = tr (sad[Xi,Zi]) i=1 u X = tr ([s, ad(Xi)]ad(Zi)) i=1 u X = tr ([ad0(s))ad(Xi)]ad(Zi)), i=1

where ad0 is the adjoint representation of gl(n, C). Since ad(Y ) = s + n, we have ad0(ad(Y )) = ad0(s) + ad0(n). Since n is nilpotent, so is ad0(n). Also [ad0(s), ad0(n)] = ad0[s, n] = 0, so ad0(s) and ad0(n) commute.

n n m Choose a basis for C compatible with the decomposition C = ⊕i=1Vi, and let Eij denote the matrix with a 1 in the i-th row, j-th column and 0’s elsewhere. Then the set of Eij’s is a basis for gl(n, C), and compute directly that ad0(s)Eij = sEij − Eijs = λiEij − λjEij = (λi − λj)Eij. Consequently ad0(s) is diagonalizable. Thus we have shown that ad0(s) + ad0(n) is the Jordan decomposition for ad0(ad(Y )).

It follows that there exists a polynomial q without constant term, such that ad0(s) = q(ad0(ad(Y ))). If we choose a polynomial which ﬁxes 0 and maps each λi − λj onto λi − λj and compose it with q, we get a polynomial r such that ad0(s) = r(ad0(adY )), and r(0) = 0.

u X Applying this observation to the formula tr (sad(Y )) = tr ([ad0(s))ad(Xi)]ad(Zi)) which we com- i=1 puted above, we see that tr (sad(Y )) is a linear combination of terms of the form

k tr ([(ad0(ad(Y ))) ad(Xi)]adZi), with k > 0.

But by examination, each term above is tr (adW adZi) = B(W, Zi), for some W ∈ [g, g] ⊆ rad B. Thus by our hypothesis, each term above is 0, and we compute that tr (sad(Y )) = 0. This is a contradiction m X 2 because we already computed tr (sad(Y )) = |λi| > 0, so ad(Y ) must have been nilpotent in the i=1 ﬁrst place. This concludes the proof. Corollary 7.68. For any Lie algebra g, rad B ⊆ rad g. 52 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Proof. Since rad B is an ideal in g, it suﬃces to check that rad B is solvable. Let B0 denote the Killing form on rad B. Write g = rad B ⊕ s for some vector subspace s and ﬁx a basis which respects this decomposition. Since rad B is an ideal, for any X ∈ g, the matrix ad(X) has the form ∗ ∗ ad(X) = . 0 0

From this observation, compute directly that B0(X,Y ) = B(X,Y ) = 0 for all X,Y ∈ rad B. There- fore rad B is solvable by Cartan’s criterion (Theorem 7.67). Theorem 7.69 (Cartan’s Criterion for Semisimplicity). A ﬁnite-dimensional Lie algebra g is semisimple if and only if the Killing form for g is nondegenerate. Proof. If B is degenerate then rad g ⊇ rad B 6= {0}, whence g is not semisimple.

Conversely, assume g is not semisimple, so rad g 6= {0}. Let ` be the least integer for which (rad g)` = {0}. Then a = (rad g)`−1 is a nonzero abelian ideal in g.

Write g = a ⊕ s for some vector subspace s, and ﬁx a basis respecting this decomposition. For X ∈ a and Y ∈ g we have 0 ∗ ∗ ∗ ad(X) = and ad(Y ) = . 0 0 0 ∗

Then ad(X)ad(Y ) has 0’s on the diagonal, and B(X,Y ) = 0. Consequently a ⊆ rad B and B is degenerate.

Theorem 7.70. A ﬁnite-dimensional Lie algebra g is semisimple if and only if g = g1 ⊕ ... ⊕ gm, where each gi is an ideal in g which is a simple Lie algebra. In this case the decomposition is unique, and the only ideals in g are the direct sums of the gi’s.

Proof. First assume that g = g1 ⊕ ... ⊕ gm as above. Let Pi : g → gi be the projection maps. Note that if Y ∈ g, Xi ∈ gi, then Y = Y1 + ... + Ym for Yi ∈ gi, and consequently Pi[Y,Xi] = Pi([Y1,Xi] + ... + [Ym,Xi]) = [Yi,Xi] = [PiY,Xi].

Suppose a is an ideal in g. Set ai = P (a). Then ai is an ideal in gi, since [PiA, Xi] = Pi[A, Xi] ⊆ Piai for all A ∈ a, Xi ∈ gi by the above remarks. Since gi is simple, we have either ai is trivial or ai = gi. In the former case, we have nothing to show. In the latter case, we have

gi = [gi, gi]

= [gi,Pi(a)]

= [gi, a] ⊆ [g, a] ⊆ a.

Consequently, a = ⊕gi⊆agi. This proves the uniqueness and the structure of the ideals. Moreover compute directly that [a, a] = a, and hence a cannot be solvable unless it is trivial. This shows g is semisimple.

Conversely, assume g is semisimple. Let a be a minimal nontrivial ideal. Let B denote the Killing form on g and let B0 denote the Killing form on a. Deﬁne a⊥ = {X ∈ g : B(X,Y ) = 0 for all Y ∈ a}.

Observe that a⊥ is an ideal in g by the invariance of the Killing form. Consequently a∩a⊥ is an ideal, ⊥ which means it is either {0} or all of a by the minimality of a. However, we also have a ∩ a = rad B0. Thus if it equals all of a, we get that a is solvable by Cartan’s criterion (Theorem 7.67), contradicting LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 53

the semisimplicity of g. So a ∩ a⊥ = {0}.

Now consider the dual space g∗ consisting of all linear mappings from g into R, which satisﬁes dim g∗ = dim g. Deﬁne a linear mapping ψ : g → g∗ by ψ(X)Y = B(X,Y ) for X,Y ∈ g.

Note that ker ψ = rad B, which is trivial since g is semisimple by Cartan’s criterion (Theorem 7.69). So ψ is an isomorphism of g onto g∗.

∗ By restricting each ψ(X) to a, we also obtain a mapping ψa : g → a deﬁned by the rule

ψa(X)A = B(X,A) for X ∈ g, A ∈ a.

⊥ ∗ We have a = ker ψa. Since ψ is an isomorphism of g onto g , it is easy to check that ψa is surjective onto a∗. Therefore

dim g = dim ker ψa + dim ranψa = dim a⊥ + dim a∗ = dim a⊥ + dim a.

Since a ∩ a⊥ = {0}, we conclude that g = a ⊕ a⊥ as vector spaces. If b ⊆ a is an ideal in a, then [b, a⊥] = {0} and so b is an ideal in g; thus b = a or b = {0} by minimality of a. This shows a is simple.

Similarly any ideal in a⊥ is an ideal in g, and hence rad a⊥ = {0}. So a⊥ is semisimple. Therefore we can repeat the argument above for a⊥, and after repeating the argument finitely many times, we get the decomposition stated in the theorem. Corollary 7.71. Let g be a semisimple Lie algebra. Then [g, g] = g. Proof. It is clear from the definition of simplicity that a simple Lie algebra h satisfies h = [h, h]. For g semisimple, write g = g1 ⊕ ... ⊕ gn with each gi simple, and compute directly to prove the corollary. 8. Haar Measure 8.1. Existence and Uniqueness of Haar Measures. Definition 8.1. A topological group is locally compact if it admits a neighborhood base consisting of open sets with compact closure. Example 8.2. (1) Since Lie groups are locally Euclidean, they are locally compact. (2) Any group with the discrete topology is locally compact. (3) Q is not locally compact. Y (4) An infinite product Gi of non-compact groups Gi is not locally compact. Neither is an i∈N infinite-dimensional Banach space. Lemma 8.3. Let G be a topological group, K ⊆ G compact, and U ⊆ G open with K ⊆ U. Then there is an open neighborhood V of e such that KV ⊆ U.

Proof. For each x ∈ K, let Wx be an neighborhood of e such that xWx ⊆ U and let Vx be an open neigh- 2 borhood of e such that Vx ⊆ Wx. Then {xVx : x ∈ K} is an open cover of K. So choose x1, ..., xn ∈ K [ such that K ⊆ xiVxi . i≤n

Set V = Vx1 ∩ · · · ∩ Vxn . If x ∈ K, then there is an i such that x ∈ xiVi. Thus

xV ⊆ xiVxi Vxi ⊆ xiWxi ⊆ U 54 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Since x was arbitrary, KV ⊆ U. Deﬁnition 8.4. Let X be a set. A σ-algebra of subsets of X is a family A of subsets of X with the properties that (1) X, ∅ ∈ A. (2) if A ∈ A then X\A ∈ A. [ \ (3) if Ai ∈ A for each i ∈ N, then Ai ∈ A and Ai ∈ A. i∈N i∈N Now let X be a topological space. Since an arbitrary intersection of σ-algebras is again a σ-algebra, there exists a least σ-algebra of subsets of X which contains all the open subsets of X. Denote this B(X), the σ-algebra of Borel subsets of X.

A function µ : B(X) → [0, ∞] is called a Borel measure on X if the following conditions hold: (1) µ(∅) = 0. (2) If A, B ∈ B(X) and A ⊆ B then µ(A) ≤ µ(B). ! ∞ [ X (3) If Ai ∈ B(X) for each i ∈ N and Ai ∩ Aj = ∅ whenever i 6= j, then µ Ai = µ(Ai). i∈N i=0 A Borel measure µ on X is regular if (1) If K ⊆ X is compact, then µ(K) < ∞. (2) If U ⊆ X is open, then µ(U) = sup{µ(K): K ⊆ U is compact}. This condition is called inner regularity. (3) If A ⊆ X is Borel, then µ(A) = inf{µ(U): U ⊇ A is open}. This condition is called outer regularity.

Now let G be a topological group. A Borel measure on G is called left translation-invariant if µ(gA) = µ(A), for every g ∈ G and A ∈ B(G). Similarly one may deﬁne a right translation-invariant measure on G.

A nonzero left translation-invariant regular Borel measure on G is called a left Haar measure. (Similarly define a right Haar measure.) Theorem 8.5. Let G be a locally compact Hausdorff topological group. Then a left Haar measure on G exists. Moreover, if µ and ν are both left Haar measures on G, then µ = cν for some c ∈ R. Proof. Let K(G) be the set of all compact subsets of G, and let U be all open neighborhoods of e. For K ∈ K(G), define

#(K : V ) = the least n such that there are x1, ··· , xn ∈ G with K ⊆ x1V ∪ · · · ∪ xnV .

◦ By the local compactness of G, ﬁx now and forever a compact K0 ⊆ G with K0 6= ∅. For each U ∈ U, we deﬁne hU : K(G) → R by

#(K:U) hU (K) = . #(K0:U)

Claim. Let U ∈ U. Then hU has the following properties:

(1) #(K : U) ≤ #(K : K0)#(K0 : U). (2) 0 ≤ hU (K) ≤ #(K : K0). (3) hU (K0) = 1. (4) hU (xK) = hU (K). (5) If K ⊆ L, then hU (K) ≤ hU (L). (6) hU (K ∪ L) ≤ hU (K) + hU (L) −1 −1 (7) If KU ∩ LU = ∅, then hU (K ∪ L) = hU (K) + hU (L). LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 55

Proof of Claim. (2) follows from (1), and everything else other than (7) is clear from the deﬁnition of −1 −1 hU . So suppose that KU ∩ LU = ∅. Let n = #(K ∪ L : U) and let x1, ..., xn ∈ G such that K ∪ L ⊆ x1U ∪ ... ∪ xnU. We wish to show that each xiU meets either K or L, but not both. Suppose toward a contradiction that there is an i such that xiU meets both K and L. Let y ∈ xiU ∩ K. Then −1 −1 there is a u ∈ U such that y = xiu. So xi = yu and thus xi ∈ KU . Similarly, if z ∈ xiU ∩ L, then −1 −1 −1 xi ∈ LU . Thus xi ∈ KU ∩ LU , a contradiction.

Now, informally speaking, we now wish to take a “limit” of the hU ’s as U shrinks to {e}. For each Q K ∈ K(G), set IK = [0, #(K : K0)] ⊆ R. Set X = K∈K(G) IK . Then by Tychonoff’s theorem, X is compact and for each U ∈ U, hU ∈ X. For each V ∈ U, set S(V ) = {hU : U ⊆ V }. Now suppose T that V1, ..., Vn ∈ U. Let V = V1 ∩ ... ∩ Vn. Then V ∈ U, so hV ∈ i≤n S(Vi). Thus the collection {S(V ): V ∈ U} of closed sets has the finite intersection property. Therefore, as X is compact, we can T fix an h ∈ V ∈U S(V ). Claim. h has the following properties: (1) 0 ≤ h(K). (2) h(∅) = 0. (3) h(K0) = 1. (4) h(xK) = h(K). (5) If K ⊆ L, then h(K) ≤ h(L). (6) h(K ∪ L) ≤ h(K) + h(L). (7) If K ∩ L = ∅, then h(K ∪ L) = h(K) + h(L).

Proof of Claim. The crucial fact for this proof is that the evaluation maps EK : X → R, given by EK (f) = f(K), are continuous since X has the product topology. We present some of the parts, but leave the rest as exercise.

−1 Let K ⊆ G be compact. Since EK is continuous, EK ([0, ∞]) is closed. For each U ∈ U, hU ∈ −1 −1 −1 EK ([0, ∞]). Thus for each V ∈ U, S(V ) ⊆ EK ([0, ∞]). So h ∈ EK ([0, ∞]), and thus h(K) ≥ 0. Since K was arbitrary, this proves (1).

Again, let K ⊆ G be compact. Then φ = ExK − EK is continuous, where the operation here is pointwise subtraction. Then φ−1(0) is closed, and we repeat the argument in the previous paragraph to see that h(xK) = h(K). Since K was arbitrary, this proves (4).

We ﬁnish by proving (7). Suppose that K ∩ L = ∅. As G is T2, we can choose U1,U2 which separate K and L. By Lemma 8.3, we can ﬁnd V1,V2 open neighborhoods of e such that KV1 ⊆ U1 and LV2 ⊆ U2. −1 −1 Let V = V1 ∩ V2. Then V is still an open neighborhood of e. Now for any U ⊆ V , U ⊆ V , and thus KU −1 ∩ LU −1 = ∅. By part (7) of Claim 1, we then have that

hU (K) + hU (L) − hU (K ∪ L) = 0.

−1 Consider φ = EK + EL − EK∪L. Then φ is continuous, so φ (0) is a closed set containing hU for −1 −1 −1 each U ⊆ V . Thus h ∈ S(V ) ⊆ ϕ (0). Now we deﬁne an outer measure µ∗ : P(G) → [0, ∞], If U ⊆ G is open, deﬁne µ∗(U) = sup{h(K): K ⊆ U is compact}.

For A ∈ P(G), we deﬁne µ∗(A) = inf{µ∗(U): U ⊇ A is open}.

We aim to apply the Caratheodory condition to create a Borel measure. We must ﬁrst check the following: (1) µ∗ is non-negative, 56 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

(2) µ∗ is monotone, (3) µ∗(∅) = 0, (4) µ∗ is countably subadditive.

The first three conditions are clear from the definition of µ∗ and the properties of h. It remains only to confirm subadditivity. Because we have defined µ∗ to be outer regular, it suffices to check open sets. S Let U = i∈ω Ui be a countable union of open sets. Let K ⊆ U be compact. Then by a standard fact about compact sets, there are compact sets K1, ..., Kn such that K = K1 ∪ ... ∪ Kn and Ki ⊆ Ui for each i ≤ n. Then

P P ∗ P ∗ h(K) ≤ i≤n h(Ki) ≤ i≤n µ (Ui) ≤ i∈ω µ (Ui)

∗ P ∗ Since this is true for all compact K ⊆ U, µ (U) ≤ i∈ω µ (Ui).

Let µ be µ∗ restricted to the µ∗ measurable sets. It remains to show that the Borel sets of G are µ∗ measurable. It suffices to show that the open sets of G are µ∗ measurable as the µ∗ measurable sets are a σ-algebra. Moreover, since µ∗ is outer regular, it again suffices to check only the µ∗-measurability of open sets. In other words, it suffices to check that

µ∗(V ) ≥ µ∗(V ∩ U) + µ∗(V ∩ U c)

whenever U and V are open in G. To see this, first fix > 0, and use the definition of µ∗ to find a compact set K ⊆ V ∩ U for which h(K) > µ∗(V ∩ U) − . Next choose a compact set L ⊆ V ∩ Kc for which h(L) > µ∗(V ∩ Kc) − ≥ µ∗(V ∩ U c) − . Since K and L are disjoint, we have

µ∗(V ∩ U) + µ∗(V ∩ U c) − 2 < h(K) + h(L) = h(K ∪ L) ≤ µ∗(V ).

Since was chosen arbitrarily, we have the inequality we want. Thus µ is a measure on a σ-algebra of sets which includes the Borel sets of G.

We will now show that µ is a left Haar measure. We begin with regularity. It is clear from definition that µ is outer regular. Moreover if K is compact and U is open containing K, then h(K) ≤ µ(U); now taking the infimum over all such U, we see that h(K) ≤ µ(K). This inequality, together with the definition of µ, show that µ is also inner regular. So we need to check that compact sets get finite measure. Let K ⊆ G be compact. Choose U ⊆ G open such that K ⊆ U and U is compact. We can do this as G is locally compact, K is compact, and a finite union of compact sets is compact. Then

µ(K) ≤ µ(U) ≤ h(U) < ∞.

The second inequality follows from the monotonicity of h. Thus µ is regular. Now µ is a Haar measure as h is invariant under left translations. Finally µ is nonzero as 1 = h(K0) ≤ µ(K0).

It remains only to show that µ is unique up to constant multiplication, i.e. if ν is another left Haar measure, then ν = cµ. To see this, ﬁx any nonzero nonnegative continuous function g : G → R with compact support, and let f : G → R be an arbitrary continuous function with compact support. The integrals of both functions with respect to µ and ν will be ﬁnite and the integrals of g will also be nonzero. We wish to show that the following ratio holds: R fdν R fdµ = R gdν R gdµ

which will imply that R fdν = c R fµ for the constant c = R gdν/ R gdµ. Since this will hold for each f, then the Riesz representation theorem will imply the result we want. LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 57

So let’s prove that the ratio holds. First consider an arbitrary function h : G × G → R with compact support. Using Fubini’s theorem to reverse the order of integration as necessary, and the translation- invariance of µ and ν to make appropriate variable substitutions, we ﬁnd that

ZZ ZZ h(x, y)dν(y)dµ(x) = h(y−1x, y)dµ(x)dν(y) ZZ = h(y−1, xy)dν(y)dµ(x).

(We replace x with y−1x for the ﬁrst equality and y with xy for the second.) Now apply this identity to the function h(x, y) = f(x)g(yx)/ R g(tx)dν(t) and we get: f(y−1) R f(x)dµ(x) = R g(x)dµ(x) R dν(y). R g(ty−1)dν(t)

R R Thus the ratio of fdµ to gdµ depends on f and g, but not on µ, which proves the claim. 8.2. The Modular Function. Definition 8.6. Let G be a locally compact Hausdorff group and µ a left Haar measure on G. For each g ∈ G, define a map µg : B(G) → [0, ∞] by the rule

µg(A) = µ(Ag).

Then it is straightforward to verify µg is another left Haar measure on G. Thus by the uniqueness part of Theorem 8.5, there exists a constant ∆(g) ∈ R+ such that µg(A) = ∆(g)µ(A), for every A ∈ B(G).

This map ∆ : G → R+ is called the modular function of G. Proposition 8.7. Let G be a locally compact Hausdorﬀ group. (1) The modular function ∆ : G → R+ is a continuous group homomorphism (where the codomain is viewed as a multiplicative group). (2) ∆ = 1 if and only if each left Haar measure on G is also a right Haar measure. Proof. The homomorphism property follows from the deﬁnition: for g, h ∈ G and A ∈ B(G) we have

∆(gh)µ(A) = µgh(A) = µ(Agh) = ∆(h)µ(Ag) = ∆(g)∆(h)µ(A).

Now since G is a topological group, it suﬃces to show the continuity of ∆ at identity. Let > 0. Let K ⊆ G be a compact set with µ(K) > 0. By the regularity of µ, let U ⊇ K be an open set with µ(U) < µ(K)(1 + ). By Lemma 8.3, let V be an open symmetric neighborhood of identity such that KV ⊆ U. Then for g ∈ V , we have Kg ⊆ KV ⊆ U and hence ∆(g)µ(K) = µ(Kg) ≤ µ(U) < µ(K)(1+). So ∆(g) < 1 + . On the other hand since V is symmetric, we have Kg−1 ⊆ KV ⊆ U and hence −1 1 1 ∆(g )µ(K) < µ(K)(1 + ), whence ∆(g) = ∆(g−1) > 1+ . Thus ∆ is continuous.

Clearly ∆ = 1 if and only if µ(Ag) = µ(A) for every A ∈ B(G) and for every g ∈ G, which completes the proof. Corollary 8.8. Let G be a Lie group. Then the modular function ∆ : G → R+ is a smooth homomorphism. 58 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Proof. By automatic smoothness of continuous Lie group homomorphisms.

Deﬁnition 8.9. A locally compact Hausdorﬀ group is called unimodular if its modular function is identically 1.

a b Exercise 41. Let G be the two-dimensional non-abelian Lie group : a, b ∈ , a > 0 . Com- 0 1 R pute the left and right Haar measures on G.

Fact 8.10. Let G be a Lie group. Then for each g ∈ G, ∆(g) = | det Ad(g)|.

Remark 8.11. The proof of the above uses an alternative development of the Haar measure for Lie groups which we do not have time to cover in this course.

Theorem 8.12. Let G be a locally compact Hausdorﬀ group satisfying one of the following conditions: (1) G is abelian; (2) G is compact; (3) G is topologically simple; (4) G is discrete; (5) G is connected semisimple Lie; or (6) G is connected nilpotent Lie.

Then G is unimodular.

Proof. (1) is obvious.

For (2), if G is compact then ∆(G) is a compact subgroup of R+ since ∆ is a smooth homomorphism. Consequently, ∆(G) = {1} and G is unimodular.

If G is topologically simple (i.e. G has no proper nontrivial closed normal subgroups) then either ker(∆) = G or ker(∆) = {e}. In the former case G is unimodular as claimed. In the latter case, ∆ is injective and we have ∆(gh) = ∆(g)∆(h) = ∆(h)∆(g) = ∆(hg), whence gh = hg for all g, h ∈ G. So G is abelian, hence unimodular. This proves (3).

In case (4), if G is discrete then the counting measure defined by µ(A) = #A when A is a finite subset of G and µ(A) = ∞ otherwise is a regular Borel measure, and it is plainly both left- and right-invariant. So its left and right Haar measures agree and hence G is unimodular. For (5), assume G is a connected semisimple Lie group. Since ∆ : G → R+ is a smooth homomorphism, we may compute its differential d∆ : g → h, where h is the one-dimensional (abelian) Lie algebra. By Corollary 7.71, we have

d∆(g) = d∆([g, g]) = [d∆(g), d∆(g)] = {0}.

So d∆ is trivial. Consequently ∆ : G → R+ is the trivial homomorphism since G is connected. Therefore G is unimodular. Lastly, suppose (6) holds and G is a connected nilpotent Lie group, and g is the (nilpotent) Lie algebra of G. By Corollary 7.57, exp : g → G is surjective. Thus for each g ∈ G there is X ∈ g with exp X = g, and therefore applying Proposition 2.13 we have LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 59

∆(g) = | det Ad(exp X)| = | det exp(adX)| = etr (adX) = 1

by the nilpotency of the operator adX. 8.3. Hilbert’s Fifth Problem for Compact Groups. Remark 8.13 (Hilbert’s Fifth Problem). The ﬁfth problem on Hilbert’s list is a bit open to interpreta- tion, but it more or less amounts to the following question: is every locally Euclidean topological group necessarily a Lie group? In other words, if you have a group which satisﬁes the purely C0 conditions of being a topological manifold with continuous group operations, does it necessarily follow that in fact the group is a C∞ manifold with C∞ group operations?

This problem occupied many brilliant mathematicians through the first half of the twentieth century and helped guide the development of the theory of Lie groups (for instance they motivate the use of the Haar measure). The answer to the question in the above paragraph is in the affirmative and comes in the form of the following powerful theorems. The proofs are involved, and presented in a clear and efficient manner in Tao’s book. Following Tao’s presentation, we will prove the positive answer to Hilbert’s fifth problem only in the case of compact groups. Theorem 8.14 (Gleason-Montgomery-Zippin 1952). Let G be a locally Euclidean second countable topological group. Then G is a Lie group. Theorem 8.15 (Gleason-Yamabe 1953). Let G be a connected locally compact topological group. Then for every open neighborhood U of identity in G, there exists a closed compact normal subgroup K of G such that K ⊆ U, and G/K is a Lie group. Definition 8.16. Let G be a compact Hausdorff group and µ a Haar measure on G. Since G is compact, by the regularity of µ, we know µ(G) < ∞. By scaling as necessary, let us assume µ(G) = 1. Then we call µ the normalized Haar measure on G.

We denote by L2(G) the space of all Borel-measurable functions f : G → C with the property that R 2 G |f| dµ < ∞, modulo the equivalence relation which makes two functions equivalent if and only if they agree everywhere except a µ-null set. L2(G) becomes a Hilbert space under the inner product R hf, gi = G f(x)g(x)dx.

For every y ∈ G, deﬁne the translation operator τ(y): L2(G) → L2(G) by the rule τ(y)f(x) = f(y−1x). Check easily that τ(y−1) is both the inverse and adjoint of τ(y) and therefore τ(y) is a unitary operator. Thus τ : G → U(L2(G)) becomes a continuous homomorphism from G into the group U(L2(G)) of unitary operators on the (probably inﬁnite-dimensional) Hilbert space L2(G), equipped with the strong operator topology. This homomorphism τ is called the left regular representation of G.

2 Now let g ∈ L (G) and deﬁne the convolution operator Tg by the rule R −1 Tg(f)(x) = f ∗ g(x) = f(z)g(z x)dµ(z).

2 Verify that Tg(f) is well-deﬁned for each f, g ∈ L (G), and that Tg(f): G → C is continuous and 2 2 2 hence lies in L (G). Thus Tg : L (G) → L (G) is a linear operator. 60 LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES

Proposition 8.17. Let G be a compact Hausdorﬀ group with normalized Haar measure µ, and for 2 −1 g ∈ L (G) let Tg denote the associated convolution operator. If g satisﬁes g(x ) = g(x) for each x ∈ G, then Tg is self-adjoint. Proof. For any f, g ∈ L2(g) we have

Z Z −1 hTg(f), hi = ( f(z)g(z x)dµ(z))h(x)dµ(x) ZZ = f(z)g(z−1x)h(x)dµ(x)dµ(z) ZZ = f(z)g(x)h(zx)dµ(x)dµ(z) ZZ = f(z)g(x−1)h(zx)dµ(x)dµ(z) ZZ = f(z)g(x−1z)h(x)dµ(x)dµ(z) Z Z = f(z)( h(x)g(x−1z)dµ(x))dµ(z)

= hf, Tg(h)i. Definition 8.18. A linear operator T : L2(G) → L2(G) is called compact if it maps norm-bounded sets to relatively compact sets. Fact 8.19. Let G be a compact Hausdorff group with normalized Haar measure µ, and for g ∈ L2(G) let Tg denote the associated convolution operator. Then Tg is a compact operator. Theorem 8.20 (Spectral Theorem). Let T : H → H be a compact self-adjoint operator on a complex Hilbert space H. Then there exists an at most countable sequence λ0, λ1, ... of non-zero real numbers which converge to 0, such that ∞ H = ⊕i=0Vi where V0 is the kernel of T , and Vi is the λi-eigenspace of T , and each Vi is finite-dimensional. Proposition 8.21 (Weak Peter-Weyl Theorem). Let G be a compact Hausdorff group and µ the normalized Haar measure on G. Let y ∈ G with y 6= e. Then there exists a finite-dimensional subspace V of L2(G) such that τ(y)V ⊆ V , and τ(y) is not the identity on V . Proof. Suppose for a contradiction that τ(y) is the identity on every finite-dimensional τ(y)-invariant subspace of L2(G).

2 −1 Let g ∈ L (G) satisfying g(x ) = g(x), and let Tg be the associated convolution operator. Apply 2 ∞ the Spectral Theorem 8.20 to Tg and obtain a decomposition L (G) = ⊕i=0Vi as in the statement. If f ∈ Vi for i ≥ 1, compute that Tg(τ(y)f) = τ(y)Tg(f) = τ(y)λif = λiτ(y)f, and therefore τ(y)Vi ⊆ Vi. By hypothesis, τ(y) is the identity on Vi. So τ(y)Tg(f) = Tg(τ(y)f) = Tg(f), i.e. τ(y)(f ∗ g) = f ∗ g.

But it is not hard to construct f, g where this fails. For instance, let U be an open symmetric 2 neighborhood of identity in G so small that y∈ / U . Set f = g = χU , the characteristic function of U. Then f ∗ g is nonzero at e and zero at y, which gives the desired contradiction. Theorem 8.22 (Gleason-Yamabe Theorem for Compact Groups). Let G be a compact Hausdorﬀ topological group. Let U be an open neighborhood of identity in G. Then there exists a compact normal subgroup K of G such that K ⊆ U and G/K is isomorphic to a closed subgroup of GL(k, C) for some k. Proof. Let g ∈ G\U. By Proposition 8.21, there is a ﬁnite-dimensional τ(g)-invariant subspace V of L2(G) on which τ(g) is not the identity. Identify this space with Cn for some n. Then we obtain a continuous homomorphism τg : G → GL(n, C) such that τ(g) 6= I. By continuity, there exists an open LIE GROUPS FALL 2016 (COHEN) LECTURE NOTES 61

neighborhood Ug of g such that τg(h) 6= I, for all h ∈ Ug. Cover G by the Ug’s and pass to a ﬁnite subcover Ug1 , ..., Ugm . Thus we have representations τi : G → GL(ni, C) with the property that for each g ∈ G, there is i with τi 6= I.   τ1(g) 0 ... 0  0 τ2(g) ... 0  Let ρ : G → GL(n1+...+nm, C) be the direct sum representation, given by ρ(g) =  .  ...  0 0 ... τm(g) Then ρ is a continuous homomorphism, and ker ρ = {e}, whence ρ is injective. Since G is compact, ρ is a topological isomorphism with a closed subgroup of ρ : G → GL(n1 + ... + nm, C),